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Trigonometric Substitution

Say you are asked to evaluate the integral

$\int \sqrt{x^2+9} \mathrm{d}x.$

If this integral was

$\int x\sqrt{x^2+9} \mathrm{d}x,$

you would be able to evaluate it using $$u$$-substitution. All you would have to do is set $$u=x^2+9$$ and $$\mathrm{d}u=2x dx$$. Then, you could write

$\int x\sqrt{x^2+9} \mathrm{d}x=\int \dfrac{1}{2}\sqrt{u} \mathrm{d}u,$

which we know how to integrate. However, without the $$x$$ on the outside of the square root, this trick does not work. Instead, we will need to use trigonometric substitution, a technique used for integrals with these problematic forms.

Definition of Trigonometric Substitution

Trigonometric substitution is an application of the Inverse Substitution Rule used to evaluate integrals containing expressions of the form $\sqrt{x^2+a^2},\quad \sqrt{a^2-x^2},\quad \text{and} \quad \sqrt{x^2-a^2}.$ It involves replacing $$x$$ with a trigonometric function, allowing these problematic expressions to be rewritten using trigonometric identities.

Trigonometric substitution is an 'inverse' form of $$u$$-substitution for particular types of integrals. When we perform $$u$$-substitution, we use the Substitution Rule.

The Substitution Rule states that, given an integrable function $$f$$ and a differentiable function $$g$$,

$\int f(g(x))g'(x) \mathrm{d}x=\left. \int f(u)\mathrm{d}u\right|_{u=g(x)},$

where $$\left. \int f(u)\mathrm{d}u\right|_{u=g(x)}$$ means 'evaluate the integral $$\int f(u)\mathrm{d}u$$, then replace $$u$$ with $$g(x)$$.

For definite integrals, the Substitution Rule is

$\int_a^b f(g(x)) \mathrm{d}x= \int_{g(a)}^{g(b)} f(u)\mathrm{d}u.$

When we use the Substitution Rule, we replace an expression in terms of $$x$$ with a variable $$u=g(x)$$ to make our integral easier to evaluate. For more information on the Substitution Rule, see the article Integration by Substitution.

The Inverse Substitution Rule is the Substitution Rule 'done backwards.'

The Inverse Substitution Rule states that, given an integrable function $$f$$ and a differentiable, one-to-one function $$g$$,

$\int f(x)\mathrm{d}x=\left. \int f(g(\theta))g'(\theta)\mathrm{d}\theta\right|_{\theta=g^{-1}(x)}.$

Here, $$\left. \int f(g(\theta))g'(\theta)\mathrm{d}\theta\right|_{\theta=g^{-1}(x)}$$ means 'integrate $$\int f(g(\theta))g'(\theta)\mathrm{d}\theta$$, then replace $$\theta$$ with $$g^{-1}(x)$$.'

For definite integrals, the Inverse Substitution Rule is

$\int_a^b f(x)\mathrm{d}x=\left. \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(\theta))g'(\theta)\mathrm{d}\theta\right|_{\theta=g^{-1}(x)}.$

In other words, when we use the Inverse Substitution Rule, instead of replacing an expression in terms of $$x$$ with a variable $$u$$, we replace $$x$$ with a function $$g(\theta)$$.

Say we are working with the integral

$\int x\mathrm{d}x.$

We can evaluate this integral as,

$\int x\mathrm{d}x = \dfrac{1}{2}x^2+C$

We could use the Inverse Substitution Rule on this integral by setting $$x=\theta^3$$, $$\mathrm{d}x=3\theta^2\mathrm{d}\theta$$, and $$\theta=\sqrt[3]{x}$$. This would give us the integral

$\int x\mathrm{d}x = \left.\int \theta^3(3\theta^2)\mathrm{d}\theta\right|_{\theta=\sqrt[3]{x}}=\left.\int 3\theta^5\mathrm{d}\theta\right|_{\theta=\sqrt[3]{x}}.$

Evaluating this integral, we get,

$\left.\int 3\theta^5\mathrm{d}\theta\right|_{\theta=\sqrt[3]{x}}=\left.\dfrac{3}{6}\theta^6+C\right|_{\theta=\sqrt[3]{x}}=\dfrac{1}{2}\left(\sqrt[3]{x}\right)^6+C=\dfrac{1}{2}x^2+C.$

Of course, in this case, inverse substitution is not something that you would want to do. However, as this example demonstrates, the Inverse Substitution Rule will always yield the same solution as other methods of integration.

This may seem rather counterintuitive; after all, replacing $$x$$ with a function seems like it would make our integral even more complicated than it already is.

However, as we will see with trigonometric substitution, this technique can actually drastically simplify certain types of integrals.

Trigonometric substitution is the most common form of inverse substitution. It is just inverse substitution where $$x$$ is replaced with a trigonometric function.

One important thing to note when applying the Inverse Substitution Rule is that the function $$g(\theta)$$ must be one-to-one. This just means that if $$g(a)=g(b)$$, then $$a=b$$.

The function $$f(x)=x^2$$ is not one-to-one, since $$f(2)=2^2=4$$ and $$f(-2)=(-2)^2=4$$, but $$2\neq-2$$. Graphically, this means that we can draw a horizontal line that intersects the graph of $$f(x)$$ at two different points.

Fig. 1. Example of a function that is not one-to-one.

By contrast, the function $$g(x)=x^3$$ is one-to-one, since if $$a^3=b^3$$, then $$a=b$$. Graphically, this means that there is no horizontal line that intersects the graph of $$g(x)$$ at two different points.

Fig. 2. Example of a one-to-one function.

This condition is essential to remember, especially for definite integrals. Otherwise, we may run into issues when replacing $$\theta$$ with $$g^{-1}(x)$$. If $$g$$ is not one-to-one, then there exist $$a$$ and $$b$$ such that $$g(a)=g(b)=c$$, but $$a\neq b$$. In general, $$g^{-1}(c)$$ is defined as the number that we plug in to $$g$$ to get $$c$$. However, since both $$a$$ and $$b$$ can be plugged into $$g$$ to get $$c$$, we have to choose whether $$g^{-1}(c)=a$$ or $$g^{-1}(c)=b$$. If we get it wrong, then our substituted integral may not be equivalent to our original integral.

Let's say, once again, we are trying to evaluate the integral

$\int x\mathrm{d}x.$

Of course, we know that this integral is

$\int x\mathrm{d}x = \dfrac{1}{2}x^2+C.$

But to make our integral as simple as possible, we decide to make the substitution $$x=g(\theta)=0$$, $$\mathrm{d}x=0\mathrm{d}\theta$$. This function is not one-to-one anywhere. For example, it maps $$\dfrac{\sqrt{\pi}}{17}$$ and $$3$$ to $$0$$, even though $$\dfrac{\sqrt{\pi}}{17}\neq 3$$.

$\int x\mathrm{d}x = \left. \int 0\mathrm{d}\theta\right|_{\theta=g^{-1}(x)} = \left. C\right|_{\theta=g^{-1}(x)}= C,$

where C is some constant. This is different from the answer we got above. What went wrong?

The issue with this substitution is that $$g^{-1}(x)$$ doesn't actually exist, since $$g$$ is not one-to-one. If $$x$$ is not zero, then there is no number $$c$$ such that the function $$g$$ maps $$c$$ to $$x$$. If $$x$$ is zero, then $$g$$ maps every real number to $$x$$, and there is no clear way to decide what $$g^{-1}(x)$$ should be. While we can make the substitution $$x=g(\theta)=0$$, we can't undo it. In order for the Substitution Rule to work, we need to be able to undo the substitution. Otherwise, as in this example, we may get an integral that is different from the original.

Trigonometric Substitution Table

There are three general cases in which trigonometric substitution is useful, and there is a formula giving the correct substitution in each case. In general, trigonometric substitution may be useful when the integral contains an expression of the form $$\sqrt{x^2+a^2}$$, $$\sqrt{a^2-x^2}$$, and $$\sqrt{x^2-a^2}$$. The following table summarizes all the cases in which we generally use trigonometric substitution.

 Term in the integral before substitution Substitution for$$x$$ Substitution for $$dx$$ $$\theta$$ interval Term in the integral after substitution $$\sqrt{a^2-x^2}$$ $$x=a\sin(\theta)$$ $$\mathrm{d}x=a\cos(\theta)\mathrm{d}\theta$$ $$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$$ $$|a\cos(\theta)|$$ $$\sqrt{x^2+a^2}$$ $$x=a\tan(\theta)$$ $$\mathrm{d}x=a\sec^2(\theta)\mathrm{d}\theta$$ $$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$$ $$|a\sec(\theta)|$$ $$\sqrt{x^2-a^2}$$ $$x=a\sec(\theta)$$ $$\mathrm{d}x=a\sec(\theta)\tan(\theta)\mathrm{d}\theta$$ $$\left[0,\tfrac{\pi}{2}\right)$$, $$\left(\tfrac{\pi}{2},\pi\right]$$,or $$\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$ $$|a\tan(\theta)|$$

Occasionally, these substitutions are useful even when there are no square roots.

You may be wondering at this point why we are allowed to substitute trigonometric functions for $$x$$. After all, they are not typically one-to-one functions.

Fig. 3. A trigonometric function that is not one-to-one.

This is why, when we make a trigonometric substitution, we constraint the theta values that we use. While trigonometric functions are not usually one-to-one over the entire real line, they are often one-to-one over specific intervals. We list intervals over which our trigonometric substitutions are one-to-one in the “theta interval” column in the table above.

You may also be wondering about the absolute value bars in the last column. When finding indefinite integrals, we can always in practice choose our theta intervals so that we can drop the absolute value bars. However, when finding definite integrals and, particularly, when working with $$|a\tan(\theta)|$$, we must be careful. In some cases, the absolute value bars cannot be dropped at all.

Instead of or in addition to the cases listed in the table above, we can sometimes use the hyperbolic trigonometric functions $$\sinh$$, $$\cosh$$, and $$\tanh$$ functions for integrals with this form. These functions satisfy the following equations,

\begin{align} \cosh^2(x)-\sinh^2(x)&=1 \\ \frac{\mathrm{d} }{\mathrm{d} x}\sinh(x)&=\cosh(x) \\ \frac{\mathrm{d} }{\mathrm{d} x}\cosh(x)&=\sinh(x)\end{align}

In particular, we can use $$\sinh$$ instead of tangent and $$\cosh$$ instead of secant in trigonometric substitutions.

Solving Integrals Using Trig Substitution

To solve an integral using trig substitution, the following steps can be helpful.

1. Identify which of the three terms in the table above appear in the integral.
2. Rewrite the equation as necessary to get it into the correct form.
• Completing the square and $$u$$-substitution can both be useful in this step.
3. Replace $$x$$ and $$\mathrm{d}x$$ with their corresponding trigonometric substitutions. Simplify.
• When working with definite integrals, be careful with absolute value signs.
4. Integrate.
5. Write down the solution.
• If you are working with a definite integral, evaluate the integral.
• If you are working with an indefinite integral, write the solution in terms of $$x$$.

Let's take a look at this process, starting with a closer examination of the three different cases in which we generally use trigonometric substitution.

Trigonometric Substitution Formulas

Integrals with $$\sqrt{a^2-x^2}$$

If an integral contains a term of the form $$\sqrt{a^2-x^2}$$, we typically use the substitutions $$x=a\sin(\theta)$$ and $$\mathrm{d}x=a\cos(\theta)\mathrm{d}\theta$$, where $$a$$ is positive.

Here, unless bounds of integration are specified, we assume $$\theta$$ is on the interval $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$. We use these bounds because $$a\sin(\theta)$$ is one-to-one on $$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$$. It is possible to use any interval on which $$\sin(\theta)$$ is one-to-one, but, as we will see, this interval tends to be particularly convenient.

Why this substitution in particular? Well, plugging in $$a\sin(\theta)$$ for $$x$$, $$\sqrt{a^2-x^2}$$ becomes

$\sqrt{a^2-(a\sin(\theta))^2}=\sqrt{a^2-a^2\sin^2(\theta)}=\sqrt{a^2(1-\sin^2(\theta))}.$

Using the identity $$1-\sin^2(\theta)=\cos^2(\theta)$$, this expression becomes

$\sqrt{a^2\cos^2(\theta)}=|a\cos(\theta)|.$

Since $$a$$ is assumed to be positive and $$\cos(\theta)$$ is positive on the interval $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$ (this is why this interval is particularly convenient!), $$|a\cos(\theta)|=a\cos(\theta)$$. This expression is often easier to integrate than $$\sqrt{a^2-x^2}$$.

If you are doing a definite integral, be sure to check that your bounds of integration are in this range of theta values. If they are not, you may not be able to drop the absolute value bars.

Let's rewrite the following integral using a substitution of the form $$x=a\sin(\theta)$$.

$\int \dfrac{x^2}{\sqrt{9-x^2}}\mathrm{d}x.$

This integral contains an expression of the form $$\sqrt{a^2-x^2}$$ where $$a^2=9$$. Since $$a$$ must be positive, we set $$a=3$$ and make the substitution $$x=3\sin(\theta)$$. If $$x=3\sin(\theta)$$, then $$\mathrm{d}x=3\cos(\theta)\mathrm{d}\theta$$, so our integral becomes,

\begin{align} \int \dfrac{x^2}{\sqrt{9-x^2}}\mathrm{d}x &= \int \dfrac{(3\sin(\theta))^2}{\sqrt{9-(3\sin(\theta))^2}}(3\cos(\theta))\mathrm{d}\theta \\ &= \int \dfrac{27\sin^2(\theta)\cos(\theta)}{3\cos(\theta)}\mathrm{d}\theta \\ &= \int 9\sin^2(\theta)\mathrm{d}\theta\end{align}

Integrals with $$\sqrt{x^2+a^2}$$

If an integral contains a term of the form $$\sqrt{x^2+a^2}$$, we generally use the substitutions $$x=a\tan(\theta)$$ and $$\mathrm{d}x=a\sec^2(\theta)\mathrm{d}\theta$$, where $$a$$ is positive and $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$. When we use this substitution, we get the expression,

$\sqrt{x^2+a^2}=\sqrt{(a\tan(\theta))^2+a^2}=\sqrt{a^2(\tan^2(\theta)+1)}.$

Since $$\tan^2(\theta)+1=\sec^2(\theta)$$, we can write

$\sqrt{a^2(\tan^2(\theta)+1)}=\sqrt{a^2\sec^2(\theta)}=|a\sec(\theta)|.$

Since $$a$$ is positive and $$\sec(\theta)$$ is also positive on the interval $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$, $$|a\sec(\theta)|=a\sec(\theta)$$. This is frequently much easier to integrate than our original expression.

Let's rewrite the integral $$\int \sqrt{x^2+9}\mathrm{d}x$$ using a substitution of the form $$x=a\tan(\theta)$$.

This integral contains an expression of the form $$\sqrt{x^2+a^2}$$, where $$a^2=9$$. Since we want $$a$$ to be positive, we set $$a=3$$. So, the substitutions we want to make are $$x=3\tan(\theta)$$, $$\mathrm{d}x=3\sec^2(\theta)\mathrm{d}\theta$$. Plugging these substitutions into the integral, we get,

\begin{align} \int \sqrt{x^2+9}\mathrm{d}x &= \int \sqrt{9\tan^2(\theta)+9}(3\sec^2(\theta))\mathrm{d}\theta \\ &= 3\int \sqrt{9(\tan^2(\theta)+1)}(\sec^2(\theta))\mathrm{d}\theta \\ &= 9\int \sqrt{\sec^2(\theta)}(\sec^2(\theta))\mathrm{d}\theta \\ &= 9\int \sec^3(\theta)\mathrm{d}\theta \end{align}

Integrals with $$\sqrt{x^2-a^2}$$

Finally, when an integral contains an expression of the form $$\sqrt{x^2-a^2}$$, we typically use the substitutions $$x=a\sec(\theta)$$ and $$\mathrm{d}x=a\sec(\theta)\tan(\theta)\mathrm{d}\theta$$, where $$a$$ is positive.

There are no universal conventions for which values of $$\theta$$ should be used, but each of the following intervals are commonly used. If our bounds of integration are both positive, then we can let $$0\le\theta<\tfrac{\pi}{2}$$. However, if either of our bounds of integration are negative, we cannot use this interval. This is because secant is positive on this interval, so if we tried to replace $$x$$ with secant, we would be replacing a sometimes negative expression with something that can only be positive. This does not work.

Fig. 4. A possible theta interval for secant.

To solve this problem, if our bounds of integration are both negative, we can use the interval $$\tfrac{\pi}{2}<\theta\le\pi$$, since secant is negative on this integral.

Fig. 5. Another possible theta interval for secant.

If one of our bounds of integration is positive and the other is negative, we can use the values $$\theta\in\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$. If we are working with an indefinite integral and thus do not have bounds of integration, we can work with any of these sets of theta values.

Fig. 6. A third possible theta interval for secant.

For simplicity's sake, we will default to using $$\theta\in\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$, especially for indefinite integrals. Practically speaking, you can use whichever of these sets of theta values you feel most comfortable with. If you choose not to use $$\theta\in\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$, just be sure to check your bounds of integration.

Using this substitution, we get the expression

$\sqrt{x^2-a^2}=\sqrt{(a\sec(\theta))^2-a^2}=\sqrt{a^2(\sec^2(\theta)-1)}.$

Using the identity $$\sec^2(\theta)-1=\tan^2(\theta)$$, this becomes

$\sqrt{a^2(\sec^2(\theta)-1)}=\sqrt{a^2\tan^2(\theta)}=|a\tan(\theta)|.$

If $$0\le\theta<\tfrac{\pi}{2}$$ or $$\theta\in\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$, then, since $$a$$ is positive and $$\tan(\theta)$$ is also positive on both of these sets, $$|a\tan(\theta)|=a\tan(\theta)$$. Otherwise, if $$\tfrac{\pi}{2}<\theta\le\pi$$, then $$|a\tan(\theta)|=-a\tan(\theta)$$. In either case, this expression is often easier to integrate than $$\sqrt{x^2-a^2}$$.

Let's rewrite the following integral using a substitution of the form $$x=a\sec(\theta)$$.

$\int \dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x.$

In this case, we set $$a=5$$ and make the substitution $$x=5\sec(\theta)$$, $$\mathrm{d}x=5\sec(\theta)\tan(\theta)\mathrm{d}\theta$$ over the range $$\theta\in\left[0,\tfrac{\pi}{2}\right)\cup\left[\pi,\tfrac{3\pi}{2}\right)$$:

\begin{align} \int \dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x &= \int \dfrac{\sqrt{(5\sec(\theta))^2-25}}{5\sec(\theta)}5\sec(\theta)\tan(\theta)\mathrm{d}\theta \\ &= \int \sqrt{25(\sec^2(\theta)-1)}\tan(\theta)\mathrm{d}\theta \\ &= 5 \int |\tan(\theta)|\tan(\theta)\mathrm{d}\theta \\ &= 5 \int \tan^2(\theta)\mathrm{d}\theta\end{align}

Examples of Integration with Trig Substitution

Let's put all the steps above together to evaluate a few integrals using trig substitution.

Integrals with terms that are similar to, but not the same as one of the three forms given above, can often still be solved with trigonometric substitution. The trick is to use $$u$$-substitution to get the integral into a familiar form. Let's do an example combining $$u$$-substitution and trig substitution.

Let's evaluate the following integral using trigonometric substitution,

$\int \dfrac{x^2}{\sqrt{9-4x^2}}\mathrm{d}x.$

First, we must decide which substitution to use. Although the integral does not contain anything exactly like any of the three general forms listed above, it does contain the term $$\sqrt{9-4x^2}$$, which looks similar to the form $$\sqrt{a^2-x^2}$$. So, let's try to use the substitution $$x=a\sin(\theta)$$.

Our first step is to rewrite the equation so it looks more like $$\sqrt{a^2-x^2}$$. To accomplish this, we start by making a $$u$$-substitution to get rid of the $$4$$ in the $$4x^2$$ term. Setting $$u=2x$$ so $$\mathrm{d}u=2\mathrm{d}x$$ and $$\mathrm{d}x=\tfrac{1}{2}\mathrm{d}u$$, we get the integral

\begin{align} \int \dfrac{x^2}{\sqrt{9-4x^2}}\mathrm{d}x &= \int \dfrac{\left(\dfrac{u}{2}\right)^2}{\sqrt{9-4\left(\dfrac{u}{2}\right)^2}}\dfrac{1}{2}\mathrm{d}u \\ &= \dfrac{1}{8}\int \dfrac{u^2}{\sqrt{9-u^2}}\mathrm{d}u \end{align}

Now that we have a term of the form $$\sqrt{a^2-u^2}$$, we can make the substitution $$u=a\sin(\theta)$$ and $$\mathrm{d}u=a\cos(\theta)\mathrm{d}\theta$$. In this case, we want to make the substitution $$u=3\sin(\theta)$$ and $$\mathrm{d}u=3\cos(\theta)\mathrm{d}\theta$$. We found the substituted form of this integral in the section Integrals with $$a^2-x^2$$. It is:

\begin{align} \dfrac{1}{8}\int \dfrac{u^2}{\sqrt{9-u^2}}\mathrm{d}u &= \dfrac{1}{8}\int 9\sin^2(\theta)\mathrm{d}\theta \\ &= \dfrac{9}{8}\int \sin^2(\theta)\mathrm{d}\theta \end{align}

To compute this integral, we first rewrite it using the identity $$\sin^2(\theta)=\tfrac{1}{2}(1-\cos(2\theta))$$. (This identity can be derived from the double angle formula for cosine). Doing so gives the expression,

$\dfrac{9}{8}\int \sin^2(\theta)\mathrm{d}\theta = \dfrac{9}{8}\int \dfrac{1}{2}(1-\cos(2\theta))\mathrm{d}\theta = \dfrac{9}{16}\int 1-\cos(2\theta)\mathrm{d}\theta.$

We must use another $$u$$-substitution to get rid of the $$2\theta$$ term. Since we have already used the variable $$u$$, let's set $$v=2\theta$$, $$\mathrm{d}v=2\mathrm{d}\theta$$. Substituting, we get

$\dfrac{9}{16}\int 1-\cos(2\theta)\mathrm{d}\theta = \dfrac{9}{16}\int (1-\cos(v))\dfrac{1}{2}\mathrm{d}v = \dfrac{9}{32}\int 1-\cos(v)\mathrm{d}v.$

Evaluating this integral, we get that

$\dfrac{9}{32}\int 1-\cos(v)\mathrm{d}v = \dfrac{9}{32}(v-\sin(v))+C$

Since we are working with an indefinite integral, our final step is to express our result in terms of $$x$$. We did quite a few substitutions to get to this point; to summarize, we:

• Set $$u=2x$$,
• Set $$u=3\sin(\theta)$$,
• Set $$v=2\theta$$.

To rewrite the solution in terms of $$x$$, we must work backward through each of these substitutions.

First, since we set $$v=2\theta$$, we can write

$\dfrac{9}{32}(v-\sin(v))+C=\dfrac{9}{32}(2\theta-\sin(2\theta))+C,$

undoing the last substitution step.

Since we replaced $$u$$ with $$\theta$$, we must figure out how to replace $$\theta$$ with $$u$$. Recall that we set $$u=3\sin(\theta)$$, so it may be helpful to first rewrite everything in terms of $$\theta$$ using the identity $$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$:

$\dfrac{9}{32}(2\theta-\sin(2\theta))+C=\dfrac{9}{32}(2\theta-2\sin(\theta)\cos(\theta))+C$

We can immediately rewrite the $$\theta$$ and $$\sin(\theta)$$ terms in terms of $$u$$:

$\dfrac{9}{32}(2\theta-2\sin(\theta)\cos(\theta))+C=\dfrac{9}{32}\left(2\sin^{-1}\left( \dfrac{u}{3}\right)-2\dfrac{u}{3}\cos(\theta)\right)+C$

Our only problem now is rewriting $$\cos(\theta)$$ in terms of $$u$$. To accomplish this, we interpret $$\theta$$ as being an angle of a right triangle. Recall that $$\sin(\theta)$$ can be interpreted as the length of the side of the triangle opposite theta divided by the length of the hypotenuse. Since $$\sin(\theta)=\tfrac{u}{3}$$, we can interpret $$u$$ as the length of the opposite side of the triangle and 3 as the length of the hypotenuse.

Fig. 7. Trigonometric substitutions in triangles.

With this interpretation, we can immediately write that $$\cos(\theta)=\dfrac{\sqrt{9-u^2}}{3}.$$ Plugging this in to our expression, we get that

\begin{align}&\dfrac{9}{32}\left(2\sin^{-1}\left( \dfrac{u}{3}\right)-2\dfrac{u}{3}\cos(\theta)\right)+C \\&= \dfrac{9}{32}\left(2\sin^{-1}\left( \dfrac{u}{3}\right)-2\dfrac{u}{3}\dfrac{\sqrt{9-u^2}}{3}\right)+C.\end{align}

Simplifying, we get

\begin{align}&\dfrac{9}{32}\left(2\sin^{-1}\left( \dfrac{u}{3}\right)-2\dfrac{u}{3}\dfrac{\sqrt{9-u^2}}{3}\right)+C \\&= \dfrac{9}{16}\sin^{-1}\left( \dfrac{u}{3}\right)-\dfrac{1}{16}u\sqrt{9-u^2}+C\end{align}

Finally, we can rewrite the solution in terms of $$x$$, using the fact that $$u=2x$$:

$\dfrac{9}{16}\sin^{-1}\left( \dfrac{u}{3}\right)-\dfrac{1}{16}u\sqrt{9-u^2}+C = \dfrac{9}{16}\sin^{-1}\left( \dfrac{2}{3}x\right)-\dfrac{1}{8}x\sqrt{9-4x^2}+C$

Thus, $\int \dfrac{x^2}{\sqrt{9-4x^2}}\mathrm{d}x = \dfrac{9}{16}\sin^{-1}\left( \dfrac{2}{3}x\right)-\dfrac{1}{8}x\sqrt{9-4x^2}+C$

The difficulty we ran into in this example with expressing $$\cos(\theta)$$ in terms of $$u$$ is quite common. Whenever you run into a similar problem, the trick of using triangles is quite useful. The picture below shows the triangles used for each of the three commonly used trigonometric substitutions.

Fig. 8. Triangles for trigonometric substitutions, from left to right $$x=a\sin(\theta)$$, $$x=a\tan(\theta)$$ and $$x=a \\sec(\theta)$$.

Another technique that is often useful when working with trigonometric substitution is completing the square.

Let's evaluate the following integral using trigonometric substitution

$\int_0^1 \dfrac{e^x}{e^{2x}+2e^x+2}\mathrm{d}x.$

At first, it may not look like this integral can be evaluated using $$u$$-substitution. However, if we make the substitution $$u=e^x$$, $$\mathrm{d}u=e^x\mathrm{d}x$$, we get,

$\int_0^1 \dfrac{e^x}{e^{2x}+2e^x+2}\mathrm{d}x =\int_1^e \dfrac{1}{u^2+2u+2}\mathrm{d}u$

Completing the square and using the substitution $$v=u+1$$, $$\mathrm{d}v=\mathrm{d}u$$, we get:

\begin{align} \int_1^e \dfrac{1}{u^2+2u+2}\mathrm{d}u &= \int_1^e \dfrac{1}{(u^2+2u+1)+1}\mathrm{d}u \\ &= \int_1^e \dfrac{1}{(u+1)^2+1}\mathrm{d}u \\ &= \int_2^{e+1} \dfrac{1}{v^2+1}\mathrm{d}v \end{align}

At this point, even though there is no square root sign, we can try using the substitution $$v=\tan(\theta)$$ and $$\mathrm{d}v=\sec^2(\theta)\mathrm{d}\theta$$, on the interval $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$ to get:

\begin{align}\int_2^{e+1} \dfrac{1}{v^2+1}\mathrm{d}v &= \int_{\arctan(2)}^{\arctan(e+1)} \dfrac{\sec^2(\theta)}{\tan^2(\theta)+1}\mathrm{d}\theta \\ &= \int_{\arctan(2)}^{\arctan(e+1)} \dfrac{\sec^2(\theta)}{\sec^2(\theta)}\mathrm{d}\theta \\ &= \int_{\arctan(2)}^{\arctan(e+1)} 1\mathrm{d}\theta \\ &= \left.\theta \right|_{\arctan(2)}^{\arctan(e+1)} \\ &= \arctan(e+1)-\arctan(2) \\ &\approx 0.201 \end{align}

Trigonometric Substitution – Key takeaways

• Trigonometric substitution is an inverse form of $$u$$-substitution that can be used for integrals with terms of the form $$\sqrt{x^2-a^2}$$, $$\sqrt{x^2+a^2}$$, and $$\sqrt{a^2-x^2}$$.
• When using trigonometric substitution, it is important to restrict the range of theta values so that the substitution is one-to-one.
• To undo a trigonometric substitution, use right triangle identities.
• Completing the square can be a useful tool for getting integrals into the correct form for using trigonometric substitution.

Trigonometric substitution is an integration technique that involves replacing x with a trigonometric function to simplify the integral.

There are three general cases in which we can use trigonometric substitution: when the integral includes a term of the form x2-a2, when it includes a term of the form x2+a2, or when it includes a term of the form a2-x2.

We use trigonometric substitution to simplify integrals that are difficult to solve in other ways.

There are three formulas for trigonometric substitution. When there is an x^2-a^2 term, we generally use the substitution x=asec(theta). When there is an x^2+a^2 term, we generally use the substitution x=atan(theta). When there is an a^2-x^2 term, we generally use the substitution x=asin(theta).

To solve integrals using trig substitution, first rewrite the integral into one of the three familiar forms, then substitute, then integrate. If working with an indefinite integral, after integrating, replace all theta terms with x terms.

Final Trigonometric Substitution Quiz

Question

Which trig substitution should you try for this integral?

$\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x$

$x=5\sec(\theta)$

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Question

Which trig substitution should you try for this integral?

$\int\dfrac{x}{\sqrt{25-x^2}}\mathrm{d}x$

$x=\tan(5\theta)$

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Question

Which trig substitution should you try for this integral?

$\int\dfrac{x}{\sqrt{25+x^2}}\mathrm{d}x$

$x=5\sin(\theta)$

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Question

Which of the following is the result of substituting $$x = 5\sin(\theta)$$ in this integral:

$\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x$

$\int\dfrac{\sqrt{\sin^2(\theta)-25}}{\sin(\theta)}\mathrm{d}\theta$

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Question

What conditions must the function $$g$$ satisfy to be able to make the inverse substitution $$x=g(\theta)$$?

$$g$$ must be one-to-one and differentiable.

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Question

What is the Inverse Substitution Rule?

The Inverse Substitution Rule states that, given an integrable function $$f$$ and a differentiable, one-to-one function $$g$$,

$\int f(x) \mathrm{d}x=\left.\int f(g(\theta))g'(\theta) \mathrm{d}\theta\right|_{\theta=g^{-1}(x)}$

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Question

Given that $$x = 5\sin(\theta)$$, what is the value of $$\tan(\theta)$$?

$\tan(\theta)=\dfrac{x}{\sqrt{25-x^2}}$

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Question

What integral is the result of substituting $$x = 5\tan(\theta)$$ in the integral:

$\int \dfrac{x}{\sqrt{x^2+25}} \mathrm{d}x$

$\int \dfrac{5\tan(\theta)}{\sqrt{25\tan^2(\theta)+25}}5\sec^2(\theta) \mathrm{d}\theta$

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Question

Why does trigonometric substitution work even though many trigonometric functions are not one-to-one?

Trigonometric substitution still works since we can restrict the values of theta to ranges over which the trigonometric function we are using is one-to-one.

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Question

What is the result of substituting $$x=5\sec(\theta)$$ in the integral:

$\int\dfrac{\sqrt{x^2-25}}{x}\mathrm{d}x$

$\int\dfrac{\sqrt{25 \sec^2(\theta)-25}}{5 \sec(\theta)}5 \sec(\theta) \tan(\theta)\mathrm{d}\theta$

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Question

What is the most appropriate technique to solve the following integral? $\int \dfrac{1}{\sqrt{x^2+2x+2}}\mathrm{d}x$

First - Complete the square inside the square root.

Second - Do a $$u$$-substitution.

Third - Do a trigonometric substitution.

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Question

What is the most effective substitution for $$x^2+a^2$$?

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

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Question

What is the most effective substitution for $$x^2-a^2$$?

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

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Question

What is the most effective substitution for $$a^2-x^2$$?

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

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Question

Why can we use the following identity?

$|a\cos(\theta)|=a\cos(\theta)$

Because, when doing the trigonometric substitution $$x=a\sin(\theta)$$ we have $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$ which makes $$\cos(\theta)$$ positive.

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