Trigonometric substitution

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We integrate by substitution with the appropriate trigonometric function.

We have seen that the definite integral can be interpreted as giving us the signed area under the curve. Consider the integral

$\int _{-1}^1 \sqrt {1-x^2} \d x$

This integral can be interpreted as area of the following region:

Using the graph above and basic facts from geometry, compute: $\int _{-1}^1 \sqrt {1-x^2} \d x = \answer {\pi /2}$

We now know a number of integration techniques, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation! This section introduces trigonometric substitution, a method of integration that will give us a new tool in our quest to compute more antiderivatives. This technique works on the same principle as substitution. Recall the substitution formula.

Integral Substitution Formula If $g$ is differentiable on the interval $[a,b]$ and $f$ is continuous on the interval $[g(a),g(b)]$ , then $\int _a^b f(g(x)) g'(x) \d x =\int _{g(a)}^{g(b)} f(u) \d u.$

The idea behind substitution is that by changing the variable of integration from $x$ to $u$ , then in many cases we can simplify the integrand to a form that we can integrate directly. In previous sections, we used this result to go from the left integral to the right integral. But what if we tried going from right to left?

Let’s rewrite the substitution formula in a more suggestive manner for our purposes:

This is just the substitution formula again but now we are thinking of transforming from left-to-right. Initially this may seem to complicate matters. We go from a simple integral to one that looks more complicated. However in certain cases we may not know how to integrate $f(x)$ , but we may be able to integrate $f(x(\theta )) x'(\theta )$ .

We start by applying this technique to compute the area that we previously used geometry to compute.

Compute: $\int _{-1}^1 \sqrt {1-x^2} \d x$

The difficulty with this integral is that we have a square root of a difference of two terms. We can’t directly simplify such an expression. (Recall: $\sqrt {a^2-b^2} \neq \sqrt {a^2} - \sqrt {b^2}$ .) However, consider the Pythagorean identity: $\cos ^2(\theta ) + \sin ^2(\theta ) = 1$ From this we see that $\cos ^2(\theta )= \answer [given]{1-\sin ^2(\theta )}$ .

If our expression was $\sqrt {1-\sin ^{2}(\theta )}$ then we could simplify. We could write

$\sqrt {1-\sin ^{2}(\theta )}=\sqrt {\cos ^{2}(\theta )}=|\cos (\theta )|$

This suggests that it would be useful to use a substitution $x=\sin (\theta )$ .

To be more precise, we want to introduce a new variable $\theta =\arcsin (x)$ . Note that since $\theta$ is the output of the $\arcsin$ function, we must have $\frac {-\pi }{2} \leq \theta \leq \frac {\pi }{2}$ , but we will work with the $x=\sin (\theta )$ form during our substitution.

Just as in the case of substitutions that we have worked with previously, we must convert the entire integral to be in terms of the new variable $\theta$ . We must also convert $\d x$ as well as the bounds on the integral.

$\begin{align*} x &=\sin(\theta),\\ \d x &= \cos(\theta) \d\theta, \end{align*}$

Now we change our limits of integration. We have

$\begin{align*} x &= 1,\\ \sin(\theta) &= 1,\\ \theta &= \answer[given]{\pi/2}, \end{align*}$ and $\begin{align*} x &= -1,\\ \sin(\theta) &= -1,\\ \theta &= \answer[given]{-\pi/2}. \end{align*}$ Now we may transform our integral via

$\begin{align*} \int_{-1}^1\sqrt{1-x^2} \d x &= \int_{\answer[given]{-\pi/2}}^{\answer[given]{\pi/2}} \sqrt{1-\sin^2(\theta)} \cdot \cos(\theta) \d \theta \\ &= \int_{\answer[given]{-\pi/2}}^{\answer[given]{\pi/2}} \sqrt{\cos^2(\theta)}\cdot \cos(\theta) \d\theta \\ &=\int_{\answer[given]{-\pi/2}}^{\answer[given]{\pi/2}} |\cos(\theta)| \cos\theta \d\theta. \end{align*}$

Now we must do something about the absolute value. Remember that since we introduced $\theta =\arcsin (\theta )$ , this means $\theta$ lies in $[- \pi /2, \pi /2 ]$ . Since on $[-\pi /2,\pi /2]$ , the function $\cos (\theta )$ is always positivenegativezero , we can drop the absolute value, and then employ a power-reduction formula

$\cos ^2(\theta )= \frac {1+\cos (2\theta )}{2}.$

Now we have

$\begin{align*} &= \int_{-\pi/2}^{\pi/2} \cos^2(\theta) \d \theta \\ &= \int_{-\pi/2}^{\pi/2} \frac{1+\cos(2\theta)}{2} \d \theta\\ &= \frac{1}{2} \eval{\answer[given]{\theta +\frac{1}{2}\sin(2\theta)}}_{-\pi/2}^{\pi/2}\\ &=\answer[given]{\frac{\pi}{2}}. \end{align*}$

This matches the answer that we obtained previously using geometry.

As you may recall, there are two ways to use substitution. Above we transformed the limits of integration, and worked with the variable $\theta$ . However, we could have solved the problem by finishing with an antiderivative in $x$ and then evaluating from the original $x$ -bounds $-1$ to $1$ . We should familiarize ourselves with this method as well since it will be our only option if we need to do an indefinite integral using trig substitution.

Above we found the antiderivative $\int \sqrt {1-x^{2}} \d x=\int \cos ^{2}(\theta ) \d \theta =\frac {\theta }{2} +\frac {1}{4}\sin (2\theta ) + C$

However, now we must express our answer in terms of $x$ . Now since $x=\sin (\theta )$ , we know that $\theta =\arcsin (x)$ . However what about the $\sin (2\theta )$ term? You may be tempted to just stick $\theta =\arcsin (x)$ in this term but we can do better and express the antiderivative in a simpler form.

To convert our answer to a function in $x$ , we construct a right triangle.

Recall that

$\sin (\theta )=\frac { Opposite}{Hypotenuse}$

Since our substitution was $\sin (\theta )=\frac {x}{1}$ , we can think of $x$ as the length of the side opposite the angle $\theta$ and we can think of $1$ as the length of the hypotenuse of the right triangle.

We can then use the Pythagorean Theorem to find the adjacent leg of the triangle. Using $?^{2}+x^{2}=1^{2}$ and solving for $?$ we obtain that the length of the adjacent side is $\sqrt {1-x^{2}}$ .

Now we still need to express $\sin (2\theta )$ in terms of $x$ using this triangle. However the angle in our right triangle is $\theta$ , not $2\theta$ . Recall the trig identity

$\sin (2\theta )=2\sin (\theta )\cos (\theta )$

We can then write our above antiderivative as $\frac {\theta }{2} +\frac {1}{2}\sin (\theta )\cos (\theta ) + C$

We can then replace $\sin (\theta )=x$ and using our triangle and the fact that $\cos (\theta )=\frac {Adjacent}{Hypotenuse}$ we can write $\cos (\theta )=\frac {\sqrt {1-x^{2}}}{1}$ .

Hence our antiderivative can now be written in terms of $x$ :

$\begin{align*} \int \sqrt{1-x^2} \d x &= \frac{\theta}{2} +\frac{1}{2}\sin(\theta)\cos(\theta)+C \\ &= \answer[given]{\frac{\arcsin(x)}{2} + \frac{x\sqrt{1-x^2}}{2}} + C \end{align*}$

If $\int \sqrt {1-x^2} \d x = \frac {\arcsin (x)}{2} + \frac {x\sqrt {1-x^2}}{2} + C,$ then $\int _{-1}^{1}\sqrt {1-x^2} \d x = \answer {\pi /2}$

The key idea with this example was that the expression $\sqrt {1-x^{2}}$ was difficult to work with directly and using the substitution $x=\sin (\theta )$ allowed us to simplify the expression and turn the integral into a trig integral that was easier to calculate.

There are other situations where this kind of strategy is useful. Let’s look at a few more examples.

Compute: $\int \frac {x^3}{x^2+9} \d x$

Here we don’t have a square root to contend with but we do have a sum of two terms in the denominator. (Recall: It is never ok to split a fraction over a sum or difference in the denominator!) It would be nice if there was some way to simplify the denominator into just one term.

Recall one of the alternate forms of the Pythagorean identity $\tan ^2(\theta ) + 1 = \answer [given]{\sec ^2(\theta )}$ and note that multiplying the entire equation by $9$ yields $\left (\answer [given]{3\tan (\theta )}\right )^2 + 9 = (3\sec (\theta ))^2$

This suggests that if we made the substitution $x=3\tan (\theta )$ then we could use this trig identity to simplify the denominator of our integrand.

Hence we’ll make the substitution

$\begin{align*} x &= 3\tan(\theta),\\ \d x &= \answer[given]{3\sec^2(\theta)} \d \theta, \end{align*}$

More precisely, this means that we are introducting a new variable $\theta =\arctan \left (\frac {x}{3}\right )$ and then rewriting it as $x=3\arctan (\theta )$ since that is a more useful form for our given problem. Since $\theta$ is the output of the $\arctan$ function, we must have $-\pi /2 < \theta < \pi /2$ .

Now we convert our integral to be in terms of the new variable $\theta$ :

$\begin{align*} \int \frac{x^3}{x^2+9} \d x &= \int \frac{(3\tan(\theta))^3}{(3\tan(\theta))^2+9} \cdot 3\sec^2(\theta) \d \theta\\ &=\int \frac{(3\tan(\theta))^3}{(3\sec(\theta))^2} \cdot 3\sec^2(\theta) \d \theta\\ &=9\int \tan^3(\theta) \d \theta. \end{align*}$ We’ll use our expertise with trigonometric integrals $\begin{align*} 9\int \tan^3(\theta) \d \theta &= 9\int \frac{\sin^3(\theta)}{\cos^{3}(\theta)} \d \theta\\ &= 9\int \frac{\sin^2(\theta) \sin(\theta)}{\cos^{3}(\theta)} \d \theta\\ &= 9\int \frac{(1-\cos^2(\theta)) \sin(\theta)}{\cos^{3}(\theta)} \d \theta \end{align*}$ Now we do another substitution $\begin{align*} u&=\cos(\theta),\\ \d u &= \answer[given]{-\sin(\theta)}\d \theta, \end{align*}$ to find $\begin{align*} &= -9\int \frac{1-u^2}{u^3}\d u\\ &= -9\int u^{-3}-\frac{1}{u}\d u\\ &= -9\left(\answer[given]{\frac{u^{-2}}{-2}}-\ln|u|\right) + C\\ &= \frac{9}{2u^2} + 9\ln|u| + C. \end{align*}$

. Since $u = \cos (\theta )$ we have

$\frac {9}{2u^2} + 9\ln |u| + C = \frac {9}{2\cos ^2(\theta )} + 9\ln |\cos (\theta )| + C.$

Now that we have an answer in terms of $\theta$ , we must convert it back to being a function in $x$ . To convert our answer to a function in $x$ , use a reference right triangle, noting that $3\tan (\theta ) = x$ , and so $\tan (\theta ) = x/3$ . Since $\tan (\theta )=\frac {Opposite}{Adjacent}$ we can label the triangle with $x$ as the length of the opposite side and $3$ as the length of the adjacent side. Note we have already used the Pythagorean Theorem to find the length of the hypotenuse.

Thus $\cos (\theta ) = \answer [given]{\frac {3}{\sqrt {x^{2}+9}}}$

and we may write

$\begin{align*} \int \frac{x^3}{x^2+9} \d x &= \frac{9}{2\cos^2(\theta)} + 9\ln|\cos(\theta)| + C\\ &= \frac{x^{2}+9}{2} + 9 \ln\left(\frac{3}{\sqrt{x^{2}+9}}\right) + C. \end{align*}$

Compute

$\int _{4\sqrt {2}}^{8} \frac {1}{ x^{2}\sqrt {x^{2}-16}} \d x \text { where } x > 4$

Again we have a square root of a difference of two terms. However unlike the $\sqrt {1-x^{2}}$ we encountered in the first example of this section, a substitution involving $\sin (\theta )$ will not allow us to simplify the square root. Instead we rewrite the form of the Pythagorean identity used in the previous example to get

$\sec ^{2}(\theta )-1=\tan ^{2}(\theta )$

This suggests that we should use

$\begin{align*} x &= 4\sec(\theta),\\ \d x &= \answer[given]{4\sec(\theta)\tan(\theta)} \d \theta, \end{align*}$

This means we are introducting a new variable $\theta =\arcsec \left (\frac {x}{4}\right )$ . In this problem, since $x > 4$ , we must have $0 \leq \theta < \pi /2$ . In general, one must be careful about which branch of $\arcsec$ needs to be used.

Let’s also go ahead and change our limits of integration: $\begin{align*} x &= 4\sqrt{2},\\ 4 \sec(\theta) &=4\sqrt{2},\\ \cos(\theta)&=\frac{1}{\sqrt{2}}, \\ \theta &= \answer[given]{\frac{\pi}{4}}, \end{align*}$ and $\begin{align*} x &= 8,\\ 4 \sec(\theta) &= 8,\\ \cos(\theta)&=\frac{1}{2}, \\ \theta &= \answer[given]{\frac{\pi}{3}}. \end{align*}$

Now we convert our integral to be in terms of the new variable $\theta$ :

$\begin{align*} \int_{4\sqrt{2}}^{8} \frac{1}{x^{2}\sqrt{x^{2}-16}} \d x &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{ 4\sec(\theta)\tan(\theta)}{ 4^{2}\sec^{2}(\theta) \sqrt{ 16(\sec^{2}(\theta)-1) }} \d \theta \\ &=\frac{1}{16}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\tan(\theta)}{\sec(\theta) |\tan(\theta)|} \d \theta \end{align*}$

The fact that $0\leq \theta < \pi /2$ guarantees that $\tan (\theta ) \geq 0$ .

Now we can evaluate the integral:

$\begin{align*} \frac{1}{16}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\tan(\theta)}{\sec(\theta) \tan(\theta)} \d \theta &= \frac{1}{16} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \answer[given]{\cos(\theta)} \d \theta\\ &= \eval{\frac{1}{16}\answer[given]{\sin(\theta)}}_{\frac{\pi}{4}}^{\frac{\pi}{3}}= \answer[given]{\frac{\sqrt{3}-\sqrt{2}}{32}} \end{align*}$

So far we have seen that we can use trig substitutions to simplify expressions like $x^{2}-a^{2}$ , $a^{2}-x^{2}$ , and $x^{2}+a^{2}$ . However, sometimes the integral may not be in a form where it is clear that a trig substitution will help. One very useful tool is to complete the square, and then choose a trigonometric function which allows the use of a Pythagorean identity to simplify the expression.

Let’s recall how to complete the square for a general quadratic $ax^2+bx+c$ . First we divide through by $a$ .

$\begin{align*} x^{2}+\frac{b}{a}x+\frac{c}{a} &=x^{2}+\frac{b}{a}+\left(\frac{b}{2a}\right)^{2}-\left(\frac{b}{2a}\right)^{2}+\frac{c}{a} \\ &=\left(x+\frac{b}{2a}\right)^{2}+\left(\frac{c}{a}-\left(\frac{b}{2a}\right)^{2}\right) \end{align*}$

Now, consider the following example:

Compute: $\int \frac {\sqrt {x^2+6x+5}}{x+3}\d x \text { where } x \geq -1$

Initially it does not seem that a trig sub would help, but let’s complete the square on the quadratic under the square root. $\begin{align*} x^2+6x+5 &= x^2+6x+\answer[given]{9}+(5-\answer[given]{9})\\ &= (\answer[given]{x+3})^2-4 \end{align*}$ Our integral becomes

$\int \frac {\sqrt {(x+3)^2-4}}{x+3}\d x$

We want to let $(x+3)$ equal some trigonometric function that will simplify after the application of a Pythagorean identity. Again

$\tan ^2(\theta ) + 1 = \sec ^2(\theta )$ will be helpful. We multiple this equation by $4$ and rearrange to find $(\answer [given]{2\tan (\theta )})^2 = (\answer [given]{2\sec (\theta )})^2 - 4$

Thus we use the substitution

$\begin{align*} (x+3) &= 2\sec(\theta)\\ \d x &= 2\sec(\theta)\tan(\theta) \d \theta. \end{align*}$ So making this substitution we have $\begin{align*} \int &\frac{\sqrt{(x+3)^2-4}}{x+3}\d x \\ &= \int\frac{\sqrt{(\answer[given]{2\sec(\theta)})^2-4}}{\answer[given]{2\sec(\theta)}} \cdot 2\sec(\theta)\tan(\theta)\d \theta\\ &= \int\frac{\sqrt{4\tan^{2}(\theta)}}{2\sec(\theta)} \cdot 2\sec(\theta)\tan(\theta)\d \theta\\ \end{align*}$ Since $x\geq -1$ , this guarantees that $0\leq \theta \leq \pi /2$ and thus that $\tan (\theta ) \geq 0$ :

$\int 2 \tan ^2(\theta ) \d \theta = 2\int \tan ^2(\theta ) \d \theta$

We’ll again use the Pythagorean identity $\tan ^2(\theta ) + 1 = \sec ^2(\theta )$ to write

$\begin{align*} &= 2\int \sec^2(\theta) - 1 \d \theta\\ &= \answer[given]{2\left(\tan(\theta) - \theta\right)} + C \end{align*}$ Now that we have an answer in $\theta$ , we must convert it back to being a function in $x$ . Since $x+3 = 2\sec (\theta )$ we have $\cos (\theta ) = \frac {2}{x+3},$ and using a reference triangle and using the Pythagorean Theorem to find the length of the opposite side we have:

Thus $\tan (\theta ) = \answer [given]{\frac {\sqrt {(x+3)^{2}-4}}{2}}$ and we may write $\begin{align*} \int &\frac{\sqrt{(x+3)^2-4}}{x+3} \d x = 2\left(\tan(\theta) - \theta\right) + C\\ &= \sqrt{(x+3)^{2}-4} - 2\arctan\left(\frac{\sqrt{(x+3)^{2}-4}}{2}\right) + C. \end{align*}$

Thus using the technique of completing the square, we can always rewrite any quadratic in a form where a trig substitution can be used to simplify the expression.

Summary

In general, when doing a trig substitution:

If the quadratic has the form $a^2 + x^2$ after completing the square, make the substitution $u = a\tan (\theta )$ .
If the quadratic has the form $a^2 - x^2$ after completing the square, make the substitution $u = a\sin (\theta )$
If the quadratic has the form $x^2 - a^2$ after completing the square, make the substitution $u = a\sec (\theta )$

You should always remember the other integration techniques you have learned previously; they may allow you to compute the integral more easily than trig substitutions in some cases. In particular, go back and look at example 2 in this section and see if you can compute this integral using a more efficient integration technique! In some cases, though, trig substitution will be your only option.