\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Integrals - Calculus 2 Practice

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

Calculus 2 Practice Integrals

These practice problems are to prepare you for your calculus 2 integrals exam as well as your final exam in calculus 2. The techniques required for these problems are listed below. In your class, you may not have learned all of these techniques. In the hint, we state the technique required for each problem.

Integral Techniques Required

Here are the techniques required in order to evaluate these integrals.

Calculus 1 Techniques

Basic Integral Techniques

Definite Integrals

Both Fundamental Theorems

Integration by Substitution

Integration using Partial Fractions

Basic Trig and Inverse Trig Integration

Calculus 2 Techniques

Integration By Parts1

Improper Integrals

All Trig Integration Techniques2

Notes

1. Some of these solution videos show using tables instead of integration by parts. DO NOT USE TABLES (he calls it the DI method in the video) unless your instructor specifically says you can. Most instructors don't allow this technique. Also, even if your instructor allows you to use tables, it is better not to use them anyway. It is very possible that you will get an instructor next semester that will not allow you to use tables and require you to use integration by parts. So be prepared.
2. The trig integration techniques include sine/cosine and secant/tangent integration, reduction techniques and trig substitution.

These problems are meant to be different from those found in each section. However, some duplication may occur either accidently or to show a different technique or solution.

Schaum's 3,000 Solved Problems in Calculus

Practice

Unless otherwise instructed, evaluate these integrals giving your answers in exact, completely factored form.

Basic

\( \int{ \tan^5x \sec^3x ~ dx } \)

Problem Statement

\( \int{ \tan^5x \sec^3x ~ dx } \)

Final Answer

\( \displaystyle{ \frac{1}{7}\sec^7x - \frac{2}{5}\sec^5x + \frac{1}{3}\sec^3x + C } \)

Problem Statement

\( \int{ \tan^5x \sec^3x ~ dx } \)

Solution

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Final Answer

\( \displaystyle{ \frac{1}{7}\sec^7x - \frac{2}{5}\sec^5x + \frac{1}{3}\sec^3x + C } \)

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\( \int{(x+e^x)^2}~dx \)

Problem Statement

\( \int{(x+e^x)^2}~dx \)

Final Answer

\(\displaystyle{ \frac{x^3}{3} + 2xe^x - 2e^x +\frac{e^{2x}}{2} + C }\)

Problem Statement

\( \int{(x+e^x)^2}~dx \)

Solution

The instructor uses a table for integration by parts. DO NOT DO THIS unless your instructor explicitly tells you it is okay. This technique obscures how to actually do integration by parts and most instructors will not allow it. So here is how you to integrate that part of the integral correctly.

\( \int{xe^x~dx} \)

\( u = x \to du = dx \)

\( dv = e^x ~dx \to v = e^x \)

\( \int{xe^x~dx} = xe^x - \int{e^x~dx} \)

\( xe^x - e^x \)

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Final Answer

\(\displaystyle{ \frac{x^3}{3} + 2xe^x - 2e^x +\frac{e^{2x}}{2} + C }\)

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\(\displaystyle{ \int{ \frac{\cos x}{\sin^2x - 5\sin x - 6} ~dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\cos x}{\sin^2x - 5\sin x - 6} ~dx } }\)

Final Answer

\(\displaystyle{ \frac{1}{7}\ln \left| \frac{\sin x - 6}{\sin x + 1} \right| + C }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\cos x}{\sin^2x - 5\sin x - 6} ~dx } }\)

Solution

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Final Answer

\(\displaystyle{ \frac{1}{7}\ln \left| \frac{\sin x - 6}{\sin x + 1} \right| + C }\)

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\(\displaystyle{ \int{ \frac{1}{x^3+1} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{x^3+1} ~ dx } }\)

Final Answer

\(\displaystyle{ \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| \frac{1}{\sqrt{3}}\arctan((2x-1)/\sqrt{3}) + C }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{x^3+1} ~ dx } }\)

Solution

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Final Answer

\(\displaystyle{ \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| \frac{1}{\sqrt{3}}\arctan((2x-1)/\sqrt{3}) + C }\)

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\( \int{ x\sin^2x ~ dx } \)

Problem Statement

\( \int{ x\sin^2x ~ dx } \)

Final Answer

\(\displaystyle{ \frac{x^2}{4} - \frac{1}{4}x\sin(2x) - \frac{1}{8}\cos(2x) + C }\)

Problem Statement

\( \int{ x\sin^2x ~ dx } \)

Solution

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Final Answer

\(\displaystyle{ \frac{x^2}{4} - \frac{1}{4}x\sin(2x) - \frac{1}{8}\cos(2x) + C }\)

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\( \int{ \cot^5(x) ~dx } \)

Problem Statement

\( \int{ \cot^5(x) ~dx } \)

Final Answer

\(\displaystyle{ -\frac{1}{4}\csc^4x + \csc^2x + \ln|\sin x| + C }\)

Problem Statement

\( \int{ \cot^5(x) ~dx } \)

Solution

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Final Answer

\(\displaystyle{ -\frac{1}{4}\csc^4x + \csc^2x + \ln|\sin x| + C }\)

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\( \int{ \sin^3x \cos^2x ~ dx } \)

Problem Statement

\( \int{ \sin^3x \cos^2x ~ dx } \)

Final Answer

\(\displaystyle{ -\frac{1}{3}\cos^3x + \frac{1}{5}\cos^5x + C }\)

Problem Statement

\( \int{ \sin^3x \cos^2x ~ dx } \)

Solution

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Final Answer

\(\displaystyle{ -\frac{1}{3}\cos^3x + \frac{1}{5}\cos^5x + C }\)

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\(\int{\sec^3x ~dx}\)

Problem Statement

\(\int{\sec^3x ~dx}\)

Final Answer

\( (1/2)\sec x \tan x + (1/2)\ln|\sec x + \tan x| + C \)

Problem Statement

\(\int{\sec^3x ~dx}\)

Solution

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Final Answer

\( (1/2)\sec x \tan x + (1/2)\ln|\sec x + \tan x| + C \)

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\(\displaystyle{ \int{ \frac{1}{x \sqrt{9x^2-1}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{x \sqrt{9x^2-1}} ~ dx } }\)

Final Answer

\( \text{arcsec}(3x) + C \)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{x \sqrt{9x^2-1}} ~ dx } }\)

Solution

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Final Answer

\( \text{arcsec}(3x) + C \)

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\(\int{ \cos \sqrt{x} ~ dx } \)

Problem Statement

\(\int{ \cos \sqrt{x} ~ dx } \)

Solution

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\(\displaystyle{ \int{ \frac{1}{\sqrt{x-x^2}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{\sqrt{x-x^2}} ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \frac{\ln x}{\sqrt{x}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\ln x}{\sqrt{x}} ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \frac{1}{e^x + e^{-x}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{e^x + e^{-x}} ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \log_2 x ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \log_2 x ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \frac{1}{(x^2+4)^2} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{(x^2+4)^2} ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \frac{1}{\sqrt{x^2+1}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{\sqrt{x^2+1}} ~ dx } }\)

Final Answer

\( \ln|\sqrt{x^2+1} +x| + C = \sinh^{-1} x + C \)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{\sqrt{x^2+1}} ~ dx } }\)

Solution

In the video solution, he just points out that the integrand is equal to the derivative of \( \sinh^{-1} x \). This is fine if your instructor allows it and you notice this. However, if you are required to evalute the integral, using the trig substitution \( x = \tan \theta \) will give you the natural log form of the answer. Practice problem 65 on the tangent substitution page is very similar to this problem.

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Final Answer

\( \ln|\sqrt{x^2+1} +x| + C = \sinh^{-1} x + C \)

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\( \int{ \ln( x + \sqrt{x^2+1} ) ~ dx } \)

Problem Statement

\( \int{ \ln( x + \sqrt{x^2+1} ) ~ dx } \)

Solution

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\(\int{ \tanh^{-1} x ~ dx }\)

Problem Statement

\(\int{ \tanh^{-1} x ~ dx }\)

Solution

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\(\int{ \sec^6 x ~ dx }\)

Problem Statement

\(\int{ \sec^6 x ~ dx }\)

Solution

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\( \int{ \ln(\sqrt{1+x^2}) ~dx } \)

Problem Statement

Evaluate the integral \( \int{ \ln(\sqrt{1+x^2}) ~dx } \). Give your answer in simplified, factored form.

Hint

This problem requires integration by parts to solve.

Problem Statement

Evaluate the integral \( \int{ \ln(\sqrt{1+x^2}) ~dx } \). Give your answer in simplified, factored form.

Hint

This problem requires integration by parts to solve.

Solution

The first video solves the given problem. The second video solves the similar but easier problem \(\int{ \ln(1+x^2) ~ dx }\).

Integrals ForYou - 1758 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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\(\int{ \sec x \tan x ~ dx }\)

Problem Statement

\(\int{ \sec x \tan x ~ dx }\)

Solution

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\(\int{ \sqrt{x^2+4x} ~ dx }\)

Problem Statement

\(\int{ \sqrt{x^2+4x} ~ dx }\)

Solution

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\(\displaystyle{ \int{ x^3 e^{x^2} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ x^3 e^{x^2} ~ dx } }\)

Solution

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\(\int{ 2x\ln(1+x) ~ dx }\)

Problem Statement

\(\int{ 2x\ln(1+x) ~ dx }\)

Solution

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\(\int{ \arctan(x) ~ dx }\)

Problem Statement

\(\int{ \arctan(x) ~ dx }\)

Solution

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\(\displaystyle{ \int{ \frac{\sin(1/x)}{x^3} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\sin(1/x)}{x^3} ~ dx } }\)

Solution

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\(\displaystyle{ \int{ \frac{\arctan x}{x^2} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\arctan x}{x^2} ~ dx } }\)

Solution

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\(\int{ (\ln x)^2 ~ dx }\)

Problem Statement

\(\int{ (\ln x)^2 ~ dx }\)

Solution

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\(\displaystyle{ \int{ \frac{\sqrt{x^2+4}}{x^2} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{\sqrt{x^2+4}}{x^2} ~ dx } }\)

Solution

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\(\int{ e^{\sqrt{x}} ~ dx }\)

Problem Statement

\(\int{ e^{\sqrt{x}} ~ dx }\)

Solution

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\(\displaystyle{ \int_{-1}^{1}{ \sqrt{4-x^2} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int_{-1}^{1}{ \sqrt{4-x^2} ~ dx } }\)

Final Answer

\(\displaystyle{ \sqrt{3} + \frac{2\pi}{3} }\)

Problem Statement

\(\displaystyle{ \int_{-1}^{1}{ \sqrt{4-x^2} ~ dx } }\)

Solution

The way he solves this problem in the video is quite irregular and most instructors will not allow that solution on an exam since the instructions say to 'evaluate' the integral. We would have solved it using trig substitution as follows.

\(\displaystyle{ \int_{-1}^{1}{ \sqrt{4-x^2} ~ dx } }\)

\(\displaystyle{ x = \sin \theta \to dx = 2\cos \theta ~ d\theta }\)

Drop the limits of integration for now.

\(\int{ \sqrt{ 4 - 4\sin^2\theta }~ 2\cos \theta ~ d\theta }\)

\(\int{ 4\cos^2\theta ~ d\theta }\)

\( 2 \int{ 1+\cos(2\theta) ~ d\theta }\)

\(\displaystyle{ 2\theta + \sin(2\theta) }\)

\(\displaystyle{ 2\theta + 2\sin\theta\cos\theta }\)

Convert back to x's and include limits of integration.

\(\displaystyle{ \left[ 2\arcsin(x/2) + \frac{x}{2}\sqrt{4-x^2} \right]_{-1}^{1} }\)

\(\displaystyle{ [ 2\arcsin(1/2) + \sqrt{3}/2 ] - [ 2\arcsin(-1/2) - \sqrt{3}/2 ] }\)

\(\displaystyle{ \sqrt{3} + 2\arcsin(1/2) - 2\arcsin(-1/2) }\)

\(\displaystyle{ \sqrt{3} + \frac{2\pi}{3} }\)

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Final Answer

\(\displaystyle{ \sqrt{3} + \frac{2\pi}{3} }\)

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\(\displaystyle{ \int{ \frac{1}{x^2-6x+9} ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{1}{x^2-6x+9} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires substitution to solve.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{1}{x^2-6x+9} ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \frac{-1}{x-3} + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{1}{x^2-6x+9} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires substitution to solve.

Solution

Factoring the denominator, we have \(x^2-6x+9 = (x-3)^2\).

\(\displaystyle{ \int{ \frac{1}{x^2-6x+9} ~dx } }\)

\(\displaystyle{ \int{ \frac{1}{(x-3)^2} ~dx } }\)

Let \( u = x-3 \to du = dx\)

\(\displaystyle{ \int{ \frac{1}{u^2} ~ du } }\)

\(\displaystyle{ \int{ u^{-2} ~ du } }\)

\(\displaystyle{ \frac{u^{-1}}{-1} + C }\)

\(\displaystyle{ \frac{-1}{x-3} + C }\)

Final Answer

\(\displaystyle{ \frac{-1}{x-3} + C }\)

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Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac = 0\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac = 0\)

Solution

Integrals ForYou - 4288 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\). Give your answer in simplified, factored form.

Hint

This problem requires partial fraction expansion to solve.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = \ln \abs{\frac{x+1}{x+2}} + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\). Give your answer in simplified, factored form.

Hint

This problem requires partial fraction expansion to solve.

Solution

\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } }\)

\(\displaystyle{ \int{ \frac{dx}{(x+1)(x+2)} } }\)

Use partial fraction expansion.

\(\displaystyle{ \int{ \frac{1}{x+1} + \frac{-1}{x+2} dx } }\)

\(\displaystyle{ \int{ \frac{1}{x+1} dx } + \int{ \frac{-1}{x+2} dx } }\)

\(\displaystyle{ \ln \abs{x+1} - \ln \abs{x+2} + C }\)

\(\displaystyle{ \ln \abs{\frac{x+1}{x+2}} + C }\)

Final Answer

\(\displaystyle{ \int{ \frac{dx}{x^2+3x+2} } = \ln \abs{\frac{x+1}{x+2}} + C }\)

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Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac > 0\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac > 0\)

Solution

Integrals ForYou - 4292 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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Intermediate

\(\displaystyle{ \int{ \frac{x \arcsin x}{\sqrt{1-x^2}} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{x \arcsin x}{\sqrt{1-x^2}} ~ dx } }\)

Hint

This looks harder than it is. Use integration by parts and choose \( u = \arcsin x\).

Problem Statement

\(\displaystyle{ \int{ \frac{x \arcsin x}{\sqrt{1-x^2}} ~ dx } }\)

Final Answer

\( -\arcsin x \sqrt{1-x^2} + x + C \)

Problem Statement

\(\displaystyle{ \int{ \frac{x \arcsin x}{\sqrt{1-x^2}} ~ dx } }\)

Hint

This looks harder than it is. Use integration by parts and choose \( u = \arcsin x\).

Solution

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Final Answer

\( -\arcsin x \sqrt{1-x^2} + x + C \)

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\( \int{ \sqrt{ x^2+4x+13 } ~ dx } \)

Problem Statement

\( \int{ \sqrt{ x^2+4x+13 } ~ dx } \)

Solution

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\( \int{ e^{2x} \cos x ~ dx } \)

Problem Statement

\( \int{ e^{2x} \cos x ~ dx } \)

Solution

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\(\displaystyle{ \int{ x^3 \sin(2x) ~ dx } }\)

Problem Statement

\(\displaystyle{ \int{ x^3 \sin(2x) ~ dx } }\)

Solution

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\(\displaystyle{ \int_1^2{ \sqrt{x^2-1} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int_1^2{ \sqrt{x^2-1} ~ dx } }\)

Solution

The way he solves this problem in the video is shorter than I would do it but it is non-intuitive and quite strange. Here is how I would solve this.
Since we have a \( x^2-1 \) under the square root, I would use the trig substitution \(x = \sec \theta \).

\( x = \sec \theta \to dx = \sec \theta \tan \theta ~ d\theta \)

Now we convert the limits of integration from \(x\) to \(\theta\)

\( x = 1 \to 1 = \sec \theta \to \theta = 0 \)

\( x = 2 \to 2 = \sec \theta \to \theta = \pi/3 \)

Now we write the integral in terms of \(\theta\)

\(\displaystyle{ \int_0^{\pi/3}{ \sqrt{ \sec^2 \theta - 1 } \sec \theta \tan \theta ~ d\theta } }\)

\(\displaystyle{ \int_0^{\pi/3}{ \tan \theta \sec \theta \tan \theta ~ d\theta } }\)

\(\displaystyle{ \int_0^{\pi/3}{ \sec \theta \tan^2 \theta ~ d\theta } }\)

\(\displaystyle{ \int_0^{\pi/3}{ \sec \theta (\sec^2 \theta - 1) ~ d\theta } }\)

\(\displaystyle{ \int_0^{\pi/3}{ \sec^3 \theta - \sec \theta ~ d\theta } }\)

We now have two integrals.
The first integral \( \int{ \sec^3 \theta ~ d\theta } \) can be solved using the secant reduction formula. This integral is solved in practice problem 2572.
The formula for the second integral \( \int{ \sec \theta ~ d\theta } \) is given on the basic trig integration page and derived in practice problem 2334.
Using those results, we have

\(\displaystyle{ \frac{1}{2} \left[ \sec \theta \tan \theta - \ln| \sec \theta + \tan \theta | \right]_0^{\pi/3} }\)

Evaluating at the limits we end up with

\( \sqrt{3} - (1/2)\ln( 2 + \sqrt{3}) \)

Finally, we need to check that our answer is the same as his. By plugging his first answer and our answer into a calculator, we get both answers are approximately \( 1.07 \) confirming they are equal.

blackpenredpen - 1305 video solution

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\( \int{ \sinh^3 x ~ dx } \)

Problem Statement

\( \int{ \sinh^3 x ~ dx } \)

Hint

Use the same technique that you would use if the integrand was \( \sin^3 x \).

Problem Statement

\( \int{ \sinh^3 x ~ dx } \)

Hint

Use the same technique that you would use if the integrand was \( \sin^3 x \).

Solution

blackpenredpen - 1748 video solution

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\(\int{ \arcsec x ~ dx }\)

Problem Statement

\(\int{ \arcsec x ~ dx }\)

Solution

blackpenredpen - 1762 video solution

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\(\int{ \arcsin(\sqrt{x}) ~ dx }\)

Problem Statement

\(\int{ \arcsin(\sqrt{x}) ~ dx }\)

Solution

blackpenredpen - 3760 video solution

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Evaluate \(\displaystyle{ \int{ \frac{1}{x^2-x+1} ~ dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1}{x^2-x+1} ~ dx } }\)

Solution

Integrals ForYou - 4289 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac < 0\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{1}{ax^2+bx+c} ~dx } }\) for \(b^2-4ac < 0\)

Solution

Integrals ForYou - 4290 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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\(\displaystyle{ \int{ x^5 \arctan x ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x^5 \arctan x ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts and substitution.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x^5 \arctan x ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ x^5 \arctan x ~dx } }\) \(\displaystyle{ = \frac{x^6}{6}\arctan x -\frac{1}{30}x^5 + \frac{1}{18}x^3 - \frac{1}{6}x + \frac{1}{6}\arctan x + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x^5 \arctan x ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts and substitution.

Solution

Don't forget the \( + C \) at the end of your solution.

Integrals ForYou - 4297 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ x^5 \arctan x ~dx } }\) \(\displaystyle{ = \frac{x^6}{6}\arctan x -\frac{1}{30}x^5 + \frac{1}{18}x^3 - \frac{1}{6}x + \frac{1}{6}\arctan x + C }\)

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\(\displaystyle{ \int{ x \sin x \cos x ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x \sin x \cos x ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.
Start by applying the identity \( \sin(2x) = 2\sin x \cos x \)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x \sin x \cos x ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ x \sin x \cos x ~dx } }\) \(\displaystyle{ = \frac{-x}{4} \cos(2x) + \frac{1}{8} \sin(2x) + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x \sin x \cos x ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.
Start by applying the identity \( \sin(2x) = 2\sin x \cos x \)

Solution

Integrals ForYou - 4299 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ x \sin x \cos x ~dx } }\) \(\displaystyle{ = \frac{-x}{4} \cos(2x) + \frac{1}{8} \sin(2x) + C }\)

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Advanced

\(\displaystyle{ \int_0^{\pi/2}{ \frac{\sin^3 x}{\sin^3 x + \cos^3 x} ~ dx } }\)

Problem Statement

\(\displaystyle{ \int_0^{\pi/2}{ \frac{\sin^3 x}{\sin^3 x + \cos^3 x} ~ dx } }\)

Hint

Start by dividing the numerator and denominator by \(\sin^3 x\) and using the substitution \( u = \cot x \). The identity \( \csc^2 \theta = 1 + \cot^2 \theta \) will come in handy.
This is a long and very involved problem. It will take you considerable time to solve. However, if you try it, you will hone your skills and you will be better prepared for your exam.
If you don't have time, go through the solution to make sure you understand each step.

Problem Statement

\(\displaystyle{ \int_0^{\pi/2}{ \frac{\sin^3 x}{\sin^3 x + \cos^3 x} ~ dx } }\)

Final Answer

\( \pi/4 \)

Problem Statement

\(\displaystyle{ \int_0^{\pi/2}{ \frac{\sin^3 x}{\sin^3 x + \cos^3 x} ~ dx } }\)

Hint

Start by dividing the numerator and denominator by \(\sin^3 x\) and using the substitution \( u = \cot x \). The identity \( \csc^2 \theta = 1 + \cot^2 \theta \) will come in handy.
This is a long and very involved problem. It will take you considerable time to solve. However, if you try it, you will hone your skills and you will be better prepared for your exam.
If you don't have time, go through the solution to make sure you understand each step.

Solution

The video solution is not really a solution. He just says the answer is \(\pi/4\) because it is. That wasn't good enough for me, so I wrote this detailed solution.

First, I will use the hint and divide the numerator and denominator by \(\sin^3 x\).

\(\displaystyle{ \int_0^{\pi/2}{ \frac{\sin^3 x}{\sin^3 x + \cos^3 x} \frac{1/\sin^3 x}{1/\sin^3 x} ~ dx } }\)

\(\displaystyle{ \int_0^{\pi/2}{ \frac{1}{1+\cot^3 x} ~ dx } }\)

Now I will continue using the hint and do integration by substitution with \( u = \cot x \).

\( u = \cot x \to du = -\csc^2 x ~ dx \)

Now we have a cosecant term but the rest of the hint tells us what to do with that.

\( du = -(1+u^2)dx \to dx = -du/(1+u^2) \)

Now I will write the integral in terms of \(u\) and drop the limits of integration.

\(\displaystyle{ \int{ \frac{1}{1+u^3} \frac{(-du)}{1+u^2} } }\)

\(\displaystyle{ \int{ \frac{-du}{(1+u^3)(1+u^2)} } }\)

It looks like we are going to have to expand the integrand using partial fraction expansion. In preparation for that, I will factor the denominator as much as I can.

\(\displaystyle{ \int{ \frac{-du}{(1+u)(u^2-u+1)(1+u^2)} } }\)

Partial Fraction Expansion

Now I will do partial fraction expansion on the integrand.

\(\displaystyle{ \frac{-1}{(1+u)(u^2-u+1)(1+u^2)} = \frac{A}{1+u} + \frac{Bu+C}{u^2-u+1} + \frac{Du+E}{1+u^2} }\)

\( -1 = A(u^2-u+1)(1+u^2) + (Bu+C)(1+u)(1+u^2) + (Du+E)(1+u)(u^2-u+1) \)

Now I do the following substitutions for \(u\) to get 5 equations with 5 unknowns.

\( u = -1 \to A = -1/6 \)

\( u = 0 \to 6C+6E = -5 \)

\( u = 1 \to 6B+6C+3D+3E = -1 \)

\( u = 2 \to 20B+10C+12D+6E = 1 \)

\( u = -2 \to -60B+30C-84D+42E = -29 \)

Solving these 5 equations, we get the following values for the unknowns.

\( A = -1/6, B = 2/3, C = -1/3, D = -1/2, E = -1/2 \)

Integrating

Plugging the constants into the integral and pulling out \(-1/6\) gives us this integral.

\(\displaystyle{ -\frac{1}{6} \int{ \frac{1}{1+u} -2 \frac{2u-1}{u^2-u+1} + \frac{3u+3}{1+u^2} ~ du } }\)

Now we look at each integrand term and evaluate them individually. We will leave off the constants of integration since ultimately this is a definite integral.

\(\displaystyle{ \int{ \frac{1}{1+u} ~ du } = \ln|1+u| }\)

\(\displaystyle{ \int{ -2 \frac{2u-1}{u^2-u+1} ~ du } = -2\ln|u^2-u+1| }\)

\(\displaystyle{ \int{ \frac{3u}{1+u^2} ~ du } = (3/2)\ln|1+u^2| }\)

\(\displaystyle{ \int{ \frac{3}{1+u^2} ~ du } = 3\arctan(u) }\)

Now combine the results to each integration.

\(\displaystyle{ -\frac{1}{6}\left[ \ln|1+u| - 2\ln|u^2-u+1| + (3/2)\ln|1+u^2| + 3\arctan(u) \right] }\)

Now convert back to x's using our original substitution \(u=\cot x\) and include the limits of integration.

\(\displaystyle{ -\frac{1}{6}\left[ \ln|1+\cot x| - 2\ln|\cot^2x-\cot x+1| + (3/2)\ln|1+\cot^2x| + 3\arctan(\cot x) \right]_0^{\pi/2} }\)

Evaluating At The Limits of Integration

If we try to evaluate at the limits of integration, we find that \(\cot(0) = \infty\). So before we can go any further, we need to change the form of these terms, so that we eliminate this issue. So we will replace \(\cot x\) with \( \cos x / \sin x \) and do some simplifying. Let's look at each term separately.

First Term

\(\displaystyle{ \ln| 1 + \cot x| }\)

\(\displaystyle{ \ln\left| 1 + \frac{\cos x}{\sin x} \right| }\)

\(\displaystyle{ \ln\left| \frac{\sin x + \cos x}{\sin x} \right| }\)

\(\displaystyle{ \ln|\sin x + \cos x| - \ln|\sin x| }\)

Second Term

\(\displaystyle{ \ln| \cot^2 x - \cot x + 1 | }\)

\(\displaystyle{ \ln\left| \frac{\cos^2 x}{\sin^2 x} - \frac{\cos x}{\sin x} + 1 \right| }\)

\(\displaystyle{ \ln\left| \frac{\cos^2 x - \cos x \sin x + \sin^2 x}{\sin^2 x} \right| }\)

\(\displaystyle{ \ln| 1 - \cos x \sin x | - 2\ln| \sin x | }\)

Third Term

\(\displaystyle{ \ln| 1 + \cot^2 x | }\)

\(\displaystyle{ \ln\left| 1 + \frac{\cos^2 x}{\sin^2 x} \right| }\)

\(\displaystyle{ \ln\left| \frac{\sin^2 x + \cos^2 x}{\sin^2 x} \right| }\)

\(\displaystyle{ \ln 1 - 2\ln|\sin x| = - 2\ln|\sin x| }\)

When all the terms are combined, the \( \ln|\sin x| \) all cancel, leaving this equation.
\(\displaystyle{ -\frac{1}{6}\left[ \ln|\sin x + \cos x| - 2\ln|1 - \cos x \sin x| + 3\arctan(\cot x) \right]_0^{\pi/2} }\)
After plugging in the limits of integration, all ther terms with \(\pi/2\) evaluate to zero. Indeed, all the natural log terms evaluate to zero. The only term remaining is \( 3\arctan(\cot 0) \). \( \cot 0 = +\infty \) and \( \arctan(+\infty) = \pi/2 \). What we have left is
\(\displaystyle{ \frac{1}{6}\left[ 3\frac{\pi}{2} \right] = \frac{\pi}{4} }\)

blackpenredpen - 3772 video solution

video by blackpenredpen

Final Answer

\( \pi/4 \)

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\(\displaystyle{ \int{ \frac{x \cdot e^x}{(1+x)^2} ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{x \cdot e^x}{(1+x)^2} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires integration by parts to solve.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{x \cdot e^x}{(1+x)^2} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires integration by parts to solve.

Solution

Integrals ForYou - 4276 video solution

video by Integrals ForYou

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\(\displaystyle{ \int{ \frac{1}{\cos x} ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{1}{\cos x} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires substitution and trig integration to solve. This looks like it should be easy but it isn't. There are at least three different ways to evaluate the integral, none of which may be obvious.
1. The weierstrass substitution \(t=\tan(x/2)\)
2. start with \(\int{ \sec x ~dx}\) and let \(u=\sec x + \tan x \) after multiplying by \(u/u\)
3. substitution \(u=1/\cos x + \tan x\) after multiplying the integrand by \(u/u\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \frac{1}{\cos x} ~dx } }\). Give your answer in simplified, factored form.

Hint

This problem requires substitution and trig integration to solve. This looks like it should be easy but it isn't. There are at least three different ways to evaluate the integral, none of which may be obvious.
1. The weierstrass substitution \(t=\tan(x/2)\)
2. start with \(\int{ \sec x ~dx}\) and let \(u=\sec x + \tan x \) after multiplying by \(u/u\)
3. substitution \(u=1/\cos x + \tan x\) after multiplying the integrand by \(u/u\)

Solution

This problem is solved several different ways as listed in the hint above. The last two videos are by two different instructors that both use technique 3.

Integrals ForYou - 4286 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Michel vanBiezen - 4286 video solution

video by Michel vanBiezen

Integrals ForYou - 4286 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Michel vanBiezen - 4286 video solution

video by Michel vanBiezen

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\(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\). Give your answer in simplified, factored form.

Hint

This solutions requires integration by parts.
1. Separate the integral into the sum of two integrals
2. Apply integration by parts with \(u=x\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\) \( = x e^{x^2} + C \)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\). Give your answer in simplified, factored form.

Hint

This solutions requires integration by parts.
1. Separate the integral into the sum of two integrals
2. Apply integration by parts with \(u=x\)

Solution

Integrals ForYou - 4293 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ (2x^2+1)e^{x^2} ~dx } }\) \( = x e^{x^2} + C \)

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\(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.
1. Start by applying integration by parts with \(dv = dx\)
2. Use integration by substitution with \( t = x^2 - 1 \). Note that for integration by substitution, you need to use a different variable other than \(u\) since you already used \(u\) for integration by parts.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\) \( = x \ln( x + \sqrt{x^2-1} ) - \sqrt{x^2-1} + C \)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.
1. Start by applying integration by parts with \(dv = dx\)
2. Use integration by substitution with \( t = x^2 - 1 \). Note that for integration by substitution, you need to use a different variable other than \(u\) since you already used \(u\) for integration by parts.

Solution

Integrals ForYou - 4294 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ \ln( x + \sqrt{x^2-1}) ~dx } }\) \( = x \ln( x + \sqrt{x^2-1} ) - \sqrt{x^2-1} + C \)

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\(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts, letting \(dv = dx\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\) \( = x \ln( x + \sqrt{x^2+1} ) - \sqrt{x^2+1} + C \)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts, letting \(dv = dx\)

Solution

Don't forget the \( + C \) at the end of your solution.
He also writes \(dv = 1\) when he starts integration by parts. This does not make sense. He should have written \(dv=1 \cdot dx\)

Integrals ForYou - 4298 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ \ln( x + \sqrt{x^2+1}) ~dx } }\) \( = x \ln( x + \sqrt{x^2+1} ) - \sqrt{x^2+1} + C \)

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\(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\) \(\displaystyle{ = \frac{e^{ax}}{a^2+b^2} \left[ (b\sin(bx) + a\cos(bx)) \right] + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.

Solution

Integrals ForYou - 4295 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ e^{ax} \cdot \cos(bx) ~dx } }\) \(\displaystyle{ = \frac{e^{ax}}{a^2+b^2} \left[ (b\sin(bx) + a\cos(bx)) \right] + C }\)

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\(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\). Give your answer in simplified, factored form.

Final Answer

\(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\) \(\displaystyle{ = \frac{e^{ax}}{a^2+b^2} \left[ (a\sin(bx) - b\cos(bx)) \right] + C }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\). Give your answer in simplified, factored form.

Hint

This solution requires integration by parts.

Solution

Integrals ForYou - 4296 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

Final Answer

\(\displaystyle{ \int{ e^{ax} \cdot \sin(bx) ~dx } }\) \(\displaystyle{ = \frac{e^{ax}}{a^2+b^2} \left[ (a\sin(bx) - b\cos(bx)) \right] + C }\)

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\(\displaystyle{ \int{ x (\arctan x)^2 ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x (\arctan x)^2 ~dx } }\). Give your answer in simplified, factored form.

Hint

Start with integration by parts by letting \(dv = x ~dx\). Then use integration by parts again, this time letting \(u=\arctan x\).

Problem Statement

Evaluate the integral \(\displaystyle{ \int{ x (\arctan x)^2 ~dx } }\). Give your answer in simplified, factored form.

Hint

Start with integration by parts by letting \(dv = x ~dx\). Then use integration by parts again, this time letting \(u=\arctan x\).

Solution

Integrals ForYou - 4318 video solution

Comment On Notation - Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.

video by Integrals ForYou

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Really UNDERSTAND Calculus

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Topics You Need To Understand For This Page

basics of integrals

substitution

Definite integrals are usually introduced early in the study of integration after covering the basics and integration by substitution. However, some practice problems on this page require the use of integration by parts, which is a more advanced technique usually introduced in second semester calculus. If you haven't studied integration by parts yet, no worries. You can skip those practice problems and come back later, once you have covered integration by parts.

Related Topics and Links

external links you may find helpful

WikiBooks: Infinite Sums

Wikipedia: Fundamental Theorem of Calculus (It is interesting that in wikipedia, the first and second fundamental theorems are switched, which is different than is sometimes taught in first semester calculus and discussed in Larson Calculus.)

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Practice Instructions

Unless otherwise instructed, evaluate these integrals giving your answers in exact, completely factored form.

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