## 17Calculus Integrals - Trig Substitution

### Difference Between Trig Integration and Trig Substitution

Trig integration is the evaluation of integrals that already have trig functions in the integrand.

Trig substitution, covered on this page, is a technique that takes an integrand that most likely does NOT contain any trig functions, and uses some trig identities to introduce trig functions into the integrand. Once the integral is completely transformed, then trig integration is used to evaluate the integral. Once the evaluation is complete, another set of substitutions, based on the original ones, is done to convert the result back to the original variable.

This is a great technique that introduces trig into an integral which originally doesn't have any trig. This makes integration easier and sometimes possible where it wasn't before.

If you want a full lecture on trig substitution, we recommend this video from one of our favorite lecturers.

### Prof Leonard - Calculus 2 Lecture 7.3: Integrals By Trigonometric Substitution [2hrs-9min-23secs]

video by Prof Leonard

The key to using this technique is to recognize specific sets of terms in the integrand, which will tell you the associated substitution. Table one contains a summary.

Table One - List of Factors

a is a constant; x and θ are variables

tangent

integrand

$$x^2 + a^2$$

substitution

$$x = a\tan(\theta)$$

identity

$$1+\tan^2(\theta) = \sec^2(\theta)$$

sine

integrand

$$-x^2 + a^2$$

substitution

$$x = a\sin(\theta)$$

identity

$$\sin^2(\theta) + \cos^2(\theta) = 1$$

secant

integrand

$$x^2-a^2$$

substitution

$$x = a\sec(\theta)$$

identity

$$1+\tan^2(\theta) = \sec^2(\theta)$$

Note - - Some books and instructors may use cosecant substitution instead of secant. In this case, the substitution is $$x = a\csc(\theta)$$ and the identity is $$1+\cot^2(\theta) = \csc^2(\theta)$$.

The terms you are looking for in the integrand may appear exactly as they are listed in table one or, more likely, they will be embedded in the integrand, like under a square root or with a power. They can appear in the numerator or the denominator of a fraction. Sometimes you may even have to complete the square in order to see the term in this form. In any case, once you have recognized the form of the term, you can choose the correct the substitution. Here are the steps to evaluate integrals using this technique.

1. Choose a substitution from table one depending on the term that appears in the integrand.
2. Draw a triangle based on the substitution.
3. Perform the substitution in the integral, simplify and evaluate.
4. Use the triangle from step 2 to convert your answer back to the original variable.

Definite Integrals
There is one possible variation to these steps. If your integral is a definite integral, you can follow the steps above and then finish the problem by evaluating the result at the limits of integration. Or you could convert your limits of integration using the substitution you chose in step 1 and then evaluate them after step 3 and skip step 4. This is up to you based on the problem statement, what you find the easiest to do and on what your instructor expects. If you are not sure, ask your instructor.

Setting Up Integrals Using Trig Substitution

Okay, before we get into the details of this technique, we need to know how to set up the integrals. Here is a great video showing how to set up each of the three types of problems.

### Krista King Math - Setting Up Trigonometric Substitution [13min-12secs]

video by Krista King Math

What To Do With Square Roots

So, what do you do when you have one of the factors in table one under a square root sign? Well, you do the same substitution as if the square root sign was not there. To understand this better, here is a video explaining this and showing an example.

### Dr Chris Tisdell - Integrals With Square Root Signs [15min-33secs]

video by Dr Chris Tisdell

Now, let's look at each substitution separately. The steps for each case are pretty much the same but we discuss each case separately to help you understand this technique.

Tangent Substitution

For this substitution, we also use these equations.

$$x^2 + a^2$$ integrand substitution identity

Notice that when we perform the substitution, we have

$$\begin{array}{rcl} a^2+x^2 & = & a^2 + (a\tan(\theta))^2 \\ & = & a^2 + a^2 \tan^2(\theta) \\ & = & a^2 (1 + \tan^2(\theta)) \\ & = & a^2 \sec^2(\theta) \\ \end{array}$$

In the last equation we end up with two squared terms, which we can easily take the square root of, for example, to simplify.

As mentioned in step 2 above, we need to build a triangle based on our choice of substitution, in this case $$x = a\tan(\theta) \to \tan(\theta) = x/a$$. So, we draw a right triangle, select one of the other two angles to be $$\theta$$ and label $$a$$ and $$x$$ accordingingly. Then we use the Pythagorean Theorem to get the expression for the other side. This gives us the triangle on the right.

Next, we substitute in the integral for $$x = a\tan(\theta))$$ and $$dx = a\sec^2(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Before going on to the next technique, take a few minutes to work these practice problems.

Basic Problems

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

Problem Statement

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

$$\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } = \frac{1}{2}\left( \arctan(x) + \frac{x}{x^2+1} \right) + C }$$

Problem Statement

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

Solution

### 60 video

video by Krista King Math

$$\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } = \frac{1}{2}\left( \arctan(x) + \frac{x}{x^2+1} \right) + C }$$

$$\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }$$

$$\displaystyle{ \int_{0}^{10}{\frac{dx}{x^2+25} } = \frac{1}{5}\arctan(2) }$$

Problem Statement

$$\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }$$

Solution

In this video, she makes a major mistake in notation when working definite integrals using substitution. She carries along the limits of integration for the x variable when she has an integral using θ. This is incorrect and could lead to major errors and loss of points on homework and exams. She either should drop the limits of integration while she is using θ or she should convert the limits in terms of θ. The second way is recommended since she would not have to convert the result of the integral back to x before substituting.

For converting, we use the equation $$x=5\tan(\theta)$$ and convert $$x=0 \to \theta = 0$$ and $$x=10 \to \theta = \arctan(2)$$

$$\displaystyle{\int_{0}^{10}{ \frac{5\sec^2 \theta}{(5\tan\theta)^2 + 25} d\theta }}$$    should be    $$\displaystyle{ \int_{0}^{\arctan(2)}{ \frac{5\sec^2 \theta}{(5\tan\theta)^2 + 25} d\theta } }$$

### 62 video

video by Krista King Math

$$\displaystyle{ \int_{0}^{10}{\frac{dx}{x^2+25} } = \frac{1}{5}\arctan(2) }$$

$$\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }$$

$$\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } = \ln \left|\sqrt{2}+1\right| }$$

Problem Statement

$$\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }$$

Solution

### 65 video

video by PatrickJMT

$$\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } = \ln \left|\sqrt{2}+1\right| }$$

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

Problem Statement

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

Solution

### 1826 video

video by Dr Chris Tisdell

Intermediate Problems

$$\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }$$

$$\displaystyle{ \int{ \sqrt{x^2+4} ~dx } = \frac{x\sqrt{x^2+4}}{2} + 2\ln\left| \frac{x+\sqrt{x^2+4}}{2} \right| + C }$$

Problem Statement

$$\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }$$

Solution

### 75 video

video by Krista King Math

$$\displaystyle{ \int{ \sqrt{x^2+4} ~dx } = \frac{x\sqrt{x^2+4}}{2} + 2\ln\left| \frac{x+\sqrt{x^2+4}}{2} \right| + C }$$

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }$$

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } = \frac{1}{3}(x^2-18)\sqrt{x^2+9} + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }$$

Solution

This problem is solved in two videos.

### 80 video

video by PatrickJMT

### 80 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } = \frac{1}{3}(x^2-18)\sqrt{x^2+9} + C }$$

$$\displaystyle{ \int{ \frac{dx}{a+bx^2} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{a+bx^2} } }$$

Solution

### 2210 video

video by Michel vanBiezen

Sine Substitution

$$-x^2 + a^2$$ integrand substitution identity

Using the substitution $$x = a\sin(\theta)$$ in $$a^2-x^2$$ gives us

$$\begin{array}{rcl} a^2-x^2 & = & a^2 - (a\sin(\theta))^2 \\ & = & a^2 - a^2 \sin^2(\theta) \\ & = & a^2(1-\sin^2(\theta)) \\ & = & a^2 \cos^2(\theta) \end{array}$$

This may not seem like much but think about this. If the term $$a^2-x^2$$ is under a square root, then the last term above could be simplified quite a bit.

$$\sqrt{a^2-x^2 } = \sqrt{a^2 \cos^2(\theta)} = a\cos(\theta)$$

Notice we lost the square root and now have a term that, depending on it's location in the integrand, is much easier to integrate.

As mentioned in the steps above, we need to draw a triangle based on our choice of substitution, in this case $$x = a\sin(\theta) \to \sin(\theta) = x/a$$. Using this last equation, we draw a right triangle and label the angle $$\theta$$, then sides $$x$$ and $$a$$. Then we use the Pythagorean Theorem to get the expression for the third side.

Next, we substitute in the integral for $$x = a\sin(\theta)$$ and $$dx = a\cos(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Before going on to the next technique, take a few minutes to work these practice problems.

Basic Problems

$$\displaystyle{ \int_{0}^{3}{ \sqrt{9-x^2} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{3}{ \sqrt{9-x^2} ~dx } }$$

Solution

### 1840 video

video by Dr Chris Tisdell

Intermediate Problems

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx } }$$

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx} = -(1/3)(x^2+32)\sqrt{16-x^2} + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx } }$$

Solution

His final answer is not completely factored. To finish the factoring, we would factor out a $$\displaystyle{ \frac{\sqrt{16-x^2}}{4} }$$ and combine the term inside the brackets to get a single term. Here is our work.

 $$\displaystyle{ -4^3 \left[ \frac{\sqrt{16-x^2}}{4} - \right. }$$ $$\displaystyle{ \left. \frac{1}{3} \left( \frac{\sqrt{16-x^2}}{4} \right)^3 \right] + C }$$ $$\displaystyle{ -16\sqrt{16-x^2} \left[ 1-\frac{1}{48}(16-x^2) \right] + C }$$ $$\displaystyle{ -(1/3)\sqrt{16-x^2} (x^2+32) + C}$$

### 69 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx} = -(1/3)(x^2+32)\sqrt{16-x^2} + C }$$

$$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}} ~dx } }$$

$$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}}~dx} = \frac{25}{2}\arcsin(x/5) + \frac{x}{2}\sqrt{25-x^2} + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}} ~dx } }$$

Solution

### 71 video

video by Krista King Math

$$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}}~dx} = \frac{25}{2}\arcsin(x/5) + \frac{x}{2}\sqrt{25-x^2} + C }$$

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } = 16\pi }$$

Problem Statement

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$

Solution

### 77 video

video by Dr Chris Tisdell

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } = 16\pi }$$

Secant Substitution

$$x^2-a^2$$ integrand substitution identity

In this case, we have two choices, to use secant or cosecant. Fortunately, both should work. So you may choose either based on what you are comfortable with, what your instructor requires and any other constraints. We will discuss the use of cosecant but the use of secant parallels this and we think you can easily extrapolate and alter the discussion to fit secant.

Similar to the other two cases above, we have the equation $$1+\cot^2(\theta) = \csc^2(\theta)$$ that we will use to simplify. Performing the substitution, we have

$$\begin{array}{rcl} -a^2+x^2 & = & -a^2 + (a\csc(\theta))^2 \\ & = & -a^2 + a^2 \csc^2(\theta) \\ & = & a^2(-1+\csc^2(\theta)) \\ & = & a^2 \cot^2(\theta) \end{array}$$

This is a much simpler expression, especially if the $$-a^2+x^2$$ is under a square root.

We use $$x = a~\csc(\theta) \to \csc(\theta) = x/a \to 1/\sin(\theta) = x/a \to \sin(\theta) = a/x$$   to build the triangle on the right. We choose one angle to label $$\theta$$, label the other two sides using the last equation and then calculate the expression for the third side using the Pythagorean Theorem.

Next, we substitute in the integral for $$x = a\csc(\theta)$$ and $$dx = a\csc(\theta)\cot(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Take a few minutes to work these practice problems.

Basic Problems

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }$$

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } = \ln \left| \frac{8}{5\sqrt{3}+\sqrt{59}} \right| }$$

Problem Statement

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }$$

Solution

### 63 video

video by Krista King Math

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } = \ln \left| \frac{8}{5\sqrt{3}+\sqrt{59}} \right| }$$

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }$$

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } = \ln \left| x + \sqrt{x^2-4} \right| + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }$$

Solution

### 66 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } = \ln \left| x + \sqrt{x^2-4} \right| + C }$$

Intermediate Problems

$$\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }$$

Problem Statement

$$\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }$$

$$\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } = \frac{\sqrt{3}}{32} }$$

Problem Statement

$$\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }$$

Solution

This problem is solved in three consecutive videos.

### 73 video

video by Krista King Math

### 73 video

video by Krista King Math

### 73 video

video by Krista King Math

$$\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } = \frac{\sqrt{3}}{32} }$$

$$\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }$$

$$\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } = \sqrt{x^2-25}-5\arcsec(x/5) + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }$$

Solution

### 74 video

video by Krista King Math

$$\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } = \sqrt{x^2-25}-5\arcsec(x/5) + C }$$

Before going on, here is a quick video overview of the detail just discussed. This video may help fill in any gaps in your understanding of the basics of this technique.

### Michel vanBiezen - Trig Substitution - What Is & When to Use Trig Substitution? [2min-57secs]

video by Michel vanBiezen

Expanding The Use of This Technique

In the discussion so far on this page, we have assumed that the $$x$$ variable term is always exactly $$x$$. There is a generalization of this where you might have $$[f(x)]^2 + a^2$$ and $$f(x) \neq x$$. This technique still works and you just change the formulas as follows.

Table Two

integrand

substitution

tangent

$$[f(x)]^2+a^2$$

$$f(x) = a\tan(\theta)$$

sine

$$-[f(x)]^2+a^2$$

$$f(x) = a\sin(\theta)$$

secant

$$[f(x)]^2-a^2$$

$$f(x) = a\sec(\theta)$$

f(x) can be pretty much anything (but not a constant, so check that it contains at least one variable). Here are some examples.

table three

integrand

substitution

$$e^{2x}+9$$

$$e^x = 3\tan(\theta)$$

$$-(x+2)^2+16$$

$$x+2 = 4\sin(\theta)$$

$$4x^2-25$$

$$2x = 5\sec(\theta)$$

Note: In the middle example in the last table, the integrand may not be given in this form. You may see something like $$-x^2-4x+12$$. To get $$-(x+2)^2+16$$ you would need to complete the square.

Okay, let's work some practice problems using these ideas. You don't want to work these problems until you have mastered the basic techniques in the previous sections on this page.

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } = \frac{1}{3}\arcsin(x/3) + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$

Solution

### 61 video

video by Krista King Math

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } = \frac{1}{3}\arcsin(x/3) + C }$$

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}}} = \frac{[27-2(x+2)^2](x+2)}{3[9-(x+2)^2]^{3/2}} + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$

Solution

This problem is solved in two videos.
His answer is not completely factored. To complete the problem, we would factor out $$\displaystyle{ \frac{x+2}{\sqrt{9-(x+2)^2}} }$$ and combine the remaining terms. Here is our work.

 $$\displaystyle{ \frac{x+2}{\sqrt{9-(x+2)^2}} + \frac{1}{3}\left[ \frac{x+2}{\sqrt{9-(x+2)^2}} \right]^3 + C }$$ $$\displaystyle{ \frac{x+2}{3\sqrt{9-(x+2)^2}} }$$ $$\displaystyle{ \left[ 3 + \frac{(x+2)^2}{9-(x+2)^2} \right] + C }$$ $$\displaystyle{ \frac{x+2}{3[9-(x+2)^2]^{3/2}} \left[ 3(9-(x+2)^2) + (x+2)^2 \right] + C }$$ $$\displaystyle{ \frac{(x+2)[27-2(x+2)^2]}{3[9-(x+2)^2]^{3/2}} + C }$$ It is debatable whether or not multiplying out the numerator would simplify the answer more. We will leave it factored. However, check with your instructor to see what they would expect.

### 81 video

video by PatrickJMT

### 81 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}}} = \frac{[27-2(x+2)^2](x+2)}{3[9-(x+2)^2]^{3/2}} + C }$$

$$\displaystyle{ \int_{\ln(3/2)}^{\ln 3}{ e^x \sqrt{9-e^{2x}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{\ln(3/2)}^{\ln 3}{ e^x \sqrt{9-e^{2x}} ~dx } }$$

$$\displaystyle{ \int_{ \ln(3/2)}^{\ln 3}{e^x\sqrt{9-e^{2x}} ~dx } = \frac{3}{8} \left[ 4\pi - 3\sqrt{3} \right] }$$

Problem Statement

$$\displaystyle{ \int_{\ln(3/2)}^{\ln 3}{ e^x \sqrt{9-e^{2x}} ~dx } }$$

Solution

This is an interesting problem that doesn't seem to be a candidate for trig substitution. But a simple u-substitution up-front where $$u=e^x$$ reveals how trig substitution can be used here.

Notice how each time he uses substitution, he changes the limits of integration. This is the correct technique and can help you keep from losing track of your limits or substituting the incorrect values.

### 70 video

video by PatrickJMT

$$\displaystyle{ \int_{ \ln(3/2)}^{\ln 3}{e^x\sqrt{9-e^{2x}} ~dx } = \frac{3}{8} \left[ 4\pi - 3\sqrt{3} \right] }$$

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }$$

Problem Statement

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }$$

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} = \frac{1}{2}(x+2)\sqrt{x^2+4x+5} + \frac{1}{2}\ln\left| \sqrt{x^2+4x+5}+x+2 \right| + C }$$

Problem Statement

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }$$

Solution

### 78 video

video by Krista King Math

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} = \frac{1}{2}(x+2)\sqrt{x^2+4x+5} + \frac{1}{2}\ln\left| \sqrt{x^2+4x+5}+x+2 \right| + C }$$

$$\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }$$

$$\displaystyle{ \int{ \sqrt{9+16x^2}~dx } = \frac{x\sqrt{9+16x^2}}{2} + \frac{9}{8}\ln\left| 4x + \sqrt{9+16x^2} \right| + C}$$

Problem Statement

$$\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }$$

Solution

During the course of this problem, she drops a $$9/4$$ term from her work. So the answer at the end of the video should be multiplied by $$9/4$$ to get the correct answer. [She puts a note at about the 14:22 mark that points this out.] We give the correct answer below.

### 79 video

video by Krista King Math

$$\displaystyle{ \int{ \sqrt{9+16x^2}~dx } = \frac{x\sqrt{9+16x^2}}{2} + \frac{9}{8}\ln\left| 4x + \sqrt{9+16x^2} \right| + C}$$

$$\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }$$

$$\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } = \frac{1}{3} \ln\left| \sqrt{9x^2+4}+3x \right| + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }$$

Solution

Note: The answer we show here is not the same as the answer that he shows in his video. However, they are both correct. Here is why.
His answer is $$\displaystyle{ \frac{1}{3}\ln\left| \frac{\sqrt{9x^2+4}}{2} + \frac{3x}{2}\right| + C }$$.
We simplified this as follows.

 Looking at the right side of the equation, we have a common denominator. $$\displaystyle{ \frac{1}{3}\ln\left| \frac{\sqrt{9x^2+4}}{2} + \frac{3x}{2}\right|+C}$$ $$\displaystyle{ \frac{1}{3}\ln\left| \frac{\sqrt{9x^2+4} + 3x}{2}\right|+C}$$ Next, we use the logarithm rule $$\ln(a/b) = \ln a - \ln b$$. $$\displaystyle{ \frac{1}{3} \left[ \ln\left| \sqrt{9x^2+4} + 3x \right| - \ln|2| \right] +C}$$ Now we have a constant $$-\ln(2)/3$$ that we absorb into the constant C to get our final answer. In these steps, we do what many teachers do and just absorb constants into the unknown constant $$C$$. We admit that this is not very good notation but most teachers do it. So you need to get used to it. Technically, we should say $$C_1 = C-\ln(2)/3$$ and write our answer as $$\displaystyle{\int{\frac{dx}{\sqrt{9x^2+4}}}= \frac{1}{3}\ln\left| \sqrt{9x^2+4} + 3x\right|+C_1}$$. This would make it clear that our constant $$C_1$$ is different than his constant $$C$$ in the video.

The reason we did all this is because on this website, we ask for answers to be simplified. His answer is not considered simplified.

### 64 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } = \frac{1}{3} \ln\left| \sqrt{9x^2+4}+3x \right| + C }$$

$$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }$$

$$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } = \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }$$

Solution

substitution: $$2x = 3\tan \theta$$
related equations: $$2dx = 3\sec^2 \theta ~d\theta$$    and    $$4x^2 = 9\tan^2 \theta$$

 $$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }$$ $$\displaystyle{ \int{ \frac{(3/2)\sec^2 \theta ~d\theta}{(9\tan^2 \theta + 9)^2} } }$$ $$\displaystyle{ \frac{3}{2}\int{ \frac{\sec^2 \theta ~d\theta}{81(\tan^2 \theta + 1)^2} } }$$ $$\displaystyle{\frac{1}{54} \int{ \frac{\sec^2 \theta ~d\theta}{\sec^4 \theta} } }$$ $$\displaystyle{ \frac{1}{54} \int{ \cos^2 \theta ~d\theta } }$$ $$\displaystyle{ \frac{1}{108} \int{ 1 + \cos(2\theta) ~d\theta } }$$ $$\displaystyle{ \frac{1}{108} \left[ \theta + \frac{\sin(2\theta)}{2} \right] + C }$$ $$\displaystyle{ \frac{1}{108} \left[ \theta + \sin \theta \cos \theta \right] + C }$$ $$\displaystyle{ \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }$$

$$\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } = \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }$$

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

Problem Statement

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

$$\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } = \frac{-x}{\sqrt{4x^2-1}} + C }$$

Problem Statement

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

Solution

### 72 video

video by Krista King Math

$$\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } = \frac{-x}{\sqrt{4x^2-1}} + C }$$

$$\displaystyle{ \int{ \frac{dx}{x^2+4x+1} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{x^2+4x+1} } }$$

Solution

### 1851 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }$$

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } = \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }$$

Solution

The instructor does not complete the problem in the video but does lay out how to solve it. Here is the rest of problem.

$$\displaystyle{ \frac{\sqrt{15}}{15}\int{\csc \theta ~d\theta} = \frac{-\sqrt{15}}{15}\ln\left| \csc\theta + \cot\theta \right| + C }$$
Note: This integral is solved in a practice problem on the trig integration page.
Now we need to convert back to x terms. Using the triangle from the video, we have

$$\displaystyle{ \csc\theta = \frac{1}{\sin\theta} = \frac{x-4}{\sqrt{(x-4)^2-15}} }$$

$$\displaystyle{ \cot \theta = \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{15}}{\sqrt{(x-4)^2-15}} }$$

 $$\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \csc\theta + \cot\theta \right| + C }$$ $$\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \frac{x-4}{\sqrt{(x-4)^2-15}} + \right. }$$ $$\displaystyle{ \left. \frac{\sqrt{15}}{\sqrt{(x-4)^2-15}} \right| + C }$$ $$\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \frac{x-4 + \sqrt{15}}{\sqrt{(x-4)^2-15}} \right| + C }$$

Note: This answer can be simplified more to

$$\displaystyle{ \frac{-\sqrt{15}}{30}\ln\left| \frac{ x-4 + \sqrt{15}}{x-4-\sqrt{15}} \right| + C = \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }$$

Check with your instructor to see how much simplification they require.

### 76 video

video by MIT OCW

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } = \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }$$

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }$$

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } = \frac{-1}{25} \left[ \frac{x}{\sqrt{9x^2-25}} \right] + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }$$

Solution

### 82 video

video by PatrickJMT

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } = \frac{-1}{25} \left[ \frac{x}{\sqrt{9x^2-25}} \right] + C }$$

Solve the initial value problem $$\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }$$

Problem Statement

Solve the initial value problem $$\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }$$

Hint

If you don't know how to work initial value problems, you can still follow the trig substitution part of this video.

Problem Statement

Solve the initial value problem $$\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }$$

$$\displaystyle{ y(t) = \frac{1}{15} \arctan\left(\frac{3t}{5}\right) + \frac{\pi}{60} }$$

Problem Statement

Solve the initial value problem $$\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }$$

Hint

If you don't know how to work initial value problems, you can still follow the trig substitution part of this video.

Solution

### 68 video

video by Krista King Math

$$\displaystyle{ y(t) = \frac{1}{15} \arctan\left(\frac{3t}{5}\right) + \frac{\pi}{60} }$$

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