Evaluate the integral \( \int{ x^{-3/2} ~ dx } \). Give your answer in simplified, factored form.

This problem requires only basic integration techniques to solved.

\( \int{ x^{-3/2} ~ dx } \) \(\displaystyle{ = -2x^{-1/2} + C }\)

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

This problem can be solved with only basic integration techniques. However, a second solution is given that uses substitution.

\(\displaystyle{ \int{ \frac{dx}{\sqrt{x}} } }\) \(\displaystyle{ = 2\sqrt{x} + C }\)

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

This problem can be solved using basic trig integration.

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

Evaluate the integral \( \int{ \csc^3x \sec x ~ dx } \). Give your answer in simplified, factored form.

This problem requires basic trig integration and substitution to solve.

A couple of comments are in order that will help clarify his solution.

Part way through this problem, he does a substitution, which is valid, but might not be obvious to you. He has the integral \[ \int{ \frac{1}{\sin^3 x \cos x} ~dx } \] For the next step, he uses the identity \(\sin^2 x + \cos^2 x = 1\) by replacing the one in the numerator with \(\sin^2 x + \cos^2 x\). Keep this substitution in mind as you progress through calculus. It can be handy in some cases.

For the third integral, he doesn't show the details using integration by substitution. Here they are.

\( \int{ \csc^3x \sec x ~ dx } \) \(\displaystyle{ = \ln|\tan(x)| - \frac{1}{2}\csc^2 x + C }\)

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

This problem requires integration by substitution to solve.

He doesn't show much work here to explain his solution. So here are the details.

Notice that right after he integrates, he leaves off the \(+C\). Although he adds it at the end of the problem and his answer is correct, it is incorrect to leave it off right after integrating. Make sure you don't make the same mistake. However, as usual, check with your instructor to see what they require.

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

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

This problem requires substitution to solve. This looks harder than it is. Try using the substitution \(u = 1/x^n\).

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

This problem requires basic trig integration and substitution to solve.

\(\displaystyle{ \int{ \frac{2\sin x}{\sin(2x)} ~dx } }\) \(\displaystyle{ = \ln| \sec x + \tan x | + C }\)

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

This problem requires substitution and trig integration to solve.

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

This problem can be solved using basic trig integration and substitution.

He does a simple substitution toward the end of the video clip where \(u=4x \to du = 4dx\).

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

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

This problem can be solved using basic integration techniques.

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

Evaluate the integral \(\displaystyle{ \int{ \frac{3}{x^2+4x+29} ~ dx } }\). Give your answer in simplified, factored form.

This problem can be solved using trig integration and substitution techniques.

\(\displaystyle{ \int{ \frac{3}{x^2+4x+29} ~ dx } }\) \(\displaystyle{ = \frac{3}{5} \arctan((x+2)/5) + C }\)

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

This problem can be solved using basic integration techniques. Convert the integrand to exponentials and integrate directly.

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

This problem requires substitution to solve.

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

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

Evaluate the integral \( \int{ (ax+b)^2 ~dx } \). Give your answer in simplified, factored form.

This problem requires substitution solve. You can avoid substitution by multiplying out before integrating but then you would have to factor a trinomial, which is not easy to see with the constants \(a\) and \(b\) in the polynomial. So we recommend using integration by substitution like he does in the video solution.

\(\displaystyle{ \int{ \frac{1}{(5x-2)^4} ~ dx } }\)

Evaluate the integral \( \int{ \tan(ax+b) ~ dx } \). Give your answer in simplified, factored form.

This problem requires substitution and trig integration to solve.

\(\displaystyle{ \int{ \frac{1}{x^3-4x^2} ~ dx } }\)

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

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

\(\displaystyle{ \int_3^5{ (x-3)^9 ~ dx } }\)

\(\displaystyle{ \int{ \frac{1}{\sqrt[3]{x+1}} ~ dx } }\)

\(\displaystyle{ \int{ \frac{1}{\sqrt[3]{x}+1} ~ dx } }\)

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

\(\displaystyle{ \int_0^5{ f(x) ~ dx } }\) for \( f(x) = \left\{ \begin{array}{ll} 10 & x \leq 2 \\ 3x^2-2 & x \gt 2 \end{array} \right. \)

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

\(\displaystyle{ \int{ \frac{e^{\tan x}}{1-\sin^2 x} ~ dx } }\)

In the video, he left out the work before his answer. Basically, he used integration by substitution with \( u = \tan x \).

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

Although he doesn't write his work down, he does integration by substitution with \( u = \arctan x \)

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

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

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

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

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

\(\int{ \sqrt[4]{x} ~ dx }\)

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

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

We would usually use integration by substitution and let \(u=e^x-1\) which is under the square root. However, the way he works it is to let \(u=\sqrt{e^x-1}\). It is a little more complicated at the first but the answer falls out nicely at the end when he integrates. So we suggest you do it his way.

\(\displaystyle{ 2\sqrt{e^x-1} -4\arctan\left( \frac{\sqrt{e^x-1}}{2} \right) + C }\)

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

Factor \(\sqrt{x}\) out of the denominator term.

\( 2\ln(\sqrt{x}+1) + C \)

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

To get started, factor out an \(x^2\) from the square root and combine it with the \(x^2\) outside the square root. Then move that outside term to the numerator. Now you should be able to use integration by substitution.

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

\(\displaystyle{ \int{ x^2 \sqrt[3]{1+x^3} ~ dx } }\)

\(\displaystyle{ \int_0^5{ \left\lfloor x \right\rfloor ~ dx } }\)

\( \left\lfloor x \right\rfloor \) is the floor function.
This is kind of a trick question in that there is no closed form for the integral. You need to remember that the definite integral is the area under the curve. So draw the graph of this function and calculate the area directly.

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

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

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

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

\(\int{ \tan x \ln( \cos x ) ~ dx }\)

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

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

\(\displaystyle{ \int_{1/e}^{e}{ \frac{\sqrt{1-(\ln x)^2}}{x} ~ dx } }\)

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

Start by multiplying the numerator and denominator of the fraction by \(1-x\).

\(\displaystyle{ \int{ x^{x/\ln x} ~ dx } }\)

Start by replacing the \(x\) in the base with \(e^{\ln x}\).

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

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

\(\displaystyle{ \int{ \frac{\sqrt{\tan x}}{\sin(2x)} ~ dx } }\)

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

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

Multiply the numerator and denominator by \( \csc x - \cot x \), set this factor equal to \(u\) and try integration by substitution.

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

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

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

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

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

1. Use the identity \(\sin(2x) = 2\sin x \cos x\)
2. Expand out \( (\sin x - \cos x)^2 \) and solve for \(2\sin x \cos x\). Use this in the denominator
3. Let \(u = \sin x - \cos x\) and perform integration by substitution

\(\displaystyle{ \int{ \frac{\sin x + \cos x}{\sqrt{\sin(2x)}} ~dx } }\) \( = \arcsin( \sin x - \cos x ) + C\)