## 17Calculus Integrals - Sine Substitution

##### 17Calculus

Sine Substitution

This is a great technique that introduces sine into an integral which originally doesn't have any trig. This makes integration easier and sometimes possible where it wasn't before.
Make sure you have throughly gone through the trig substitution page since this page builds on that discussion.

Once you have determined that you have an integrand containing a $$(-x^2 + a^2)$$ term, you use these equations for the substitution.

Sine Substitution

$$a$$ is a constant; $$x$$ and $$\theta$$ are variables

integrand

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

substitution

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

identity

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

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 its 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.

Okay, here are some practice problems.

Practice

Evaluate these integrals using sine substitution. Give your answers in exact, simplified and factored form.

Basic

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

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{3}{ \sqrt{9-x^2} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

Solution

### Dr Chris Tisdell - 1840 video solution

video by Dr Chris Tisdell

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Intermediate

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

$$\displaystyle{ -(1/3)(x^2+32)\sqrt{16-x^2} + C }$$

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{x^3}{\sqrt{16-x^2}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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.

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

### PatrickJMT - 69 video solution

video by PatrickJMT

$$\displaystyle{ -(1/3)(x^2+32)\sqrt{16-x^2} + C }$$

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{x^2}{\sqrt{25-x^2}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

Solution

### Krista King Math - 71 video solution

video by Krista King Math

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

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

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

Solution

### Dr Chris Tisdell - 77 video solution

video by Dr Chris Tisdell

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

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

Problem Statement

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

Solution

### Michael Penn - 4140 video solution

video by Michael Penn

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Expanding The Use of The Sine Substitution

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 substitution as follows.

Expanded Sine Substitution

integrand

substitution

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

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

Example

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

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

$$f(x)$$ can be pretty much anything except for a constant. So check that it contains at least one variable. In the example above, $$f(x)=x+2$$.

Note: In the 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 technique in the previous section on this page.

Practice

Evaluate these integrals using sine substitution. Give your answers in exact, simplified and factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

Solution

### Krista King Math - 61 video solution

video by Krista King Math

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

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

Solution

This problem is solved in two consecutive videos.
His final 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.

### PatrickJMT - 81 video solution

video by PatrickJMT

### PatrickJMT - 81 video solution

video by PatrickJMT

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

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

Problem Statement

Evaluate $$\displaystyle{ \int_{\ln(3/2)}^{\ln 3}{ e^x \sqrt{9-e^{2x}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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

Problem Statement

Evaluate $$\displaystyle{ \int_{\ln(3/2)}^{\ln 3}{ e^x \sqrt{9-e^{2x}} ~dx } }$$ using trig substitution. Give your answer in exact, simplified, factored form.

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.

### PatrickJMT - 70 video solution

video by PatrickJMT

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

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$$\displaystyle{ \int_0^2{ (2-y) \sqrt{ 9 - (y-3)^2 } ~ dy } }$$

Problem Statement

$$\displaystyle{ \int_0^2{ (2-y) \sqrt{ 9 - (y-3)^2 } ~ dy } }$$

$$\displaystyle{ \left[ \frac{-9\pi}{4} + \frac{9}{2}\arcsin(1/3) + \frac{19\sqrt{2}}{3} \right] }$$

Problem Statement

$$\displaystyle{ \int_0^2{ (2-y) \sqrt{ 9 - (y-3)^2 } ~ dy } }$$

Solution

This integral came from practice problem 3402 on the calculus 2 100 practice problems page. It is the integral (without the constants) that calculates the hydrostatic force against one side of a plate immersed in water.

 Our first inclination might be to separate this integral into two integrals by distributing the square root term across $$(2-y)$$. Although this will work, it takes longer and you end up with 3 integrals to evaluate, not just 2. So, let's simplify the term under the square root first using integration by substitution with $$u=y-3$$. $$u=y-3 \to du = dy$$ $$u=y-3 \to u+3 = y$$ When $$y=0$$, $$u=y-3 \to u = -3$$ When $$y=2$$, $$u=y-3 \to u=-1$$ Now we will rewrite the integral in terms of $$u$$. $$\displaystyle{ \int_{-3}^{-1}{ (-u-1)\sqrt{9-u^2} ~ du} }$$ To reduce the number of negative signs, we can simplify this integral. $$\displaystyle{ \int_{-1}^{-3}{ (u+1)\sqrt{9-u^2} ~ du} }$$ That last step is not necessary but losing negative signs is one of the most common errors when evaluating integrals. So that step removed $$-1$$ from $$(-u-1)$$ and absorbed into the integral by switching the limits of integration. Now is the time to distribute the square root term across $$(u+1)$$ and split the integral into two integrals. $$\displaystyle{ \int_{-1}^{-3}{ u \sqrt{9-u^2}~du } + }$$ $$\displaystyle{ \int_{-1}^{-3}{ \sqrt{9-u^2} ~du } }$$ Each integral requires different techniques. So let's evaluate them separately. The first one is the easiest, so let's start with that one. $$\displaystyle{ \int_{-1}^{-3}{ u \sqrt{9-u^2}~du } }$$ Use basic substitution with $$x=9-u^2$$. We usually use the variable $$u$$ for substitution but we already have a $$u$$ in the integral. So we chose to use $$x$$ since it had not been used in this problem yet. Otherwise we might have chosen $$v$$. $$x=9-u^2 \to dx = -2u~du$$ when $$u=-1$$, $$x=9-u^2 \to x = 8$$ when $$u=-3$$, $$x=9-u^2 \to x= 0$$ Using these substitutions, the first integral is $$\displaystyle{ \int_8^0{ x^{1/2} \frac{dx}{-2} } = }$$ $$\displaystyle{ (1/2) \int_0^8{ x^{1/2} ~dx } = }$$ $$\displaystyle{ \frac{1}{2} \left[ \frac{x^{3/2}}{3/2} \right]_0^8 = \frac{16\sqrt{2}}{3}}$$ Okay, so now let's work on the second integral. $$\displaystyle{ \int_{-1}^{-3}{ \sqrt{9-u^2} ~du } }$$ For this integral we need to use trig substitution with $$u = 3\sin\theta$$. $$u = 3\sin\theta \to du = 3\cos \theta ~ d\theta$$ Rather than changing the limits of integration in terms of $$\theta$$, we will just drop them to make sure our notation is correct and then convert back to $$u$$ before evaluating the limits of integration. Using the substitution our integral is now $$\displaystyle{ \int{ \sqrt{9-9\sin^2\theta}(3\cos \theta) ~ d\theta } }$$ $$\displaystyle{ 9 \int{ \cos^2 \theta ~ d\theta } }$$ $$\displaystyle{ 9 \int{ \frac{1+\cos(2\theta)}{2} ~ d\theta } }$$ $$\displaystyle{ (9/2) \int{ 1+\cos(2\theta) ~ d\theta } }$$ $$\displaystyle{ \frac{9}{2}\left[ \theta + \frac{\sin(2\theta)}{2} \right] }$$ Since we need to convert this result back in terms of $$u$$, use the identity $$\sin(2\theta) = 2\sin \theta \cos \theta$$ The original substitution was $$u=3\sin\theta$$. Using this we determine that $$\theta = \arcsin(u/3)$$ and $$\cos\theta = (1/3)\sqrt{9-u^2}$$. Now we have $$\displaystyle{ \frac{9}{2} \left[ \arcsin(u/3) + \frac{u}{3} \cdot \frac{\sqrt{9-u^2}}{3} \right] }$$ Okay we are making good progress. Now we need to take the last equation and evaluate it from $$-1$$ to $$-3$$. Doing this gives us the expression $$\displaystyle{ \frac{9}{2}\left[ \arcsin(-1) + (-1)(0) \right] - }$$ $$\displaystyle{ \frac{9}{2}\left[ \arcsin(-1/3) + \frac{-1}{3}\frac{\sqrt{8}}{3} \right] }$$ $$\displaystyle{ \frac{-9\pi}{4} + \frac{9}{2}\arcsin(1/3) + \sqrt{2} }$$ Okay, that last expression is the result from the second integral. Adding it to the result from the first integral will give us the final answer.

$$\displaystyle{ \left[ \frac{-9\pi}{4} + \frac{9}{2}\arcsin(1/3) + \frac{19\sqrt{2}}{3} \right] }$$

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Now, let's look at the next substitution technique, secant/cosecant substitution.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

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