# A lecture on trigonometric integrals and trigonometric substitution

## 1 Trigonometric Integrals

First, we must recall a few trigonometric identities:

 $\displaystyle\sin^{2}x+\cos^{2}x$ $\displaystyle=$ $\displaystyle 1$ (1) $\displaystyle\sec^{2}x$ $\displaystyle=$ $\displaystyle 1+\tan^{2}x$ (2) $\displaystyle\sin^{2}x$ $\displaystyle=$ $\displaystyle\frac{1-\cos(2x)}{2}$ (3) $\displaystyle\cos^{2}x$ $\displaystyle=$ $\displaystyle\frac{1+\cos(2x)}{2}$ (4) $\displaystyle\sin(2x)$ $\displaystyle=$ $\displaystyle 2\sin x\cos x$ (5) $\displaystyle\cos(2x)$ $\displaystyle=$ $\displaystyle\cos^{2}x-\sin^{2}x.$ (6)
###### Example 1.1.

$\int\sin xdx=-\cos x+C$ and $\int\cos xdx=\sin x+C$ are immediate integrals.

###### Example 1.2.

For $\int\sin^{2}xdx,\ \int\cos^{2}xdx$ we use formulas (3) and (4) respectively, e.g.

 $\int\sin^{2}xdx=\int\frac{1-\cos(2x)}{2}dx=\frac{1}{2}\int(1-\cos(2x))dx=\frac% {1}{2}\left(x-\frac{\sin(2x)}{2}\right)+C.$
###### Example 1.3.

For integrals of the form $\int\cos^{m}x\sin x\ dx$ or $\int\sin^{m}x\cos x\ dx$ we use substitution with $u=\cos x$ or $u=\sin x$ respectively, e.g.

 $\int\cos^{2}x\sin xdx=\int-u^{2}du=-\frac{u^{3}}{3}+C=-\frac{\cos^{3}x}{3}+C.% \ [u=\cos x,\ du=-\sin xdx]$

In the following examples, we use equations (1) in the forms $\sin^{2}x=1-\cos^{2}x$ or $\cos^{2}x=1-\sin^{2}x$ to transform the integral into one of the type described in Example 1.3.

###### Example 1.4.
 $\displaystyle\int\sin^{3}xdx$ $\displaystyle=$ $\displaystyle\int\sin^{2}x\sin xdx=\int(1-\cos^{2}x)\sin xdx$ $\displaystyle=$ $\displaystyle\int\sin xdx-\int\cos^{2}x\sin xdx$ $\displaystyle=$ $\displaystyle-\cos x+\frac{\cos^{3}x}{3}+C.$

Similarly one can solve $\int\cos^{3}xdx$.

###### Example 1.5.
 $\displaystyle\int\cos^{3}x\sin^{2}xdx$ $\displaystyle=$ $\displaystyle\int\cos^{2}x\cos x\sin^{2}xdx=\int(1-\sin^{2}x)\cos x\sin^{2}xdx$ $\displaystyle=$ $\displaystyle\int\cos x\sin^{2}xdx-\int\cos x\sin^{4}xdx$ $\displaystyle=$ $\displaystyle\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C.$
###### Example 1.6.

In order to solve $\int\cos^{5}x\sin^{3}xdx$ we express it first as $\int\cos^{5}x\sin^{2}x\sin x=\int\cos^{5}x(1-\cos^{2}x)\sin xdx$ and then proceed as in the previous example.

One can use similar tricks to solve integrals which involve products of powers of $\sec x$ and $\tan x$, by using Equation (2). Also, recall that the derivative of $\tan x$ is $\sec^{2}x$ while the derivative of $\sec x$ is $\sec x\tan x$.

###### Example 1.7.
 $\displaystyle\int\tan^{5}x\sec^{4}xdx$ $\displaystyle=$ $\displaystyle\int\tan^{5}x\sec^{2}x\sec^{2}xdx=\int\tan^{5}x(1+\tan^{2}x)\sec^% {2}xdx$ $\displaystyle=$ $\displaystyle\int\tan^{5}x\sec^{2}xdx+\int\tan^{7}x\sec^{2}xdx$ $\displaystyle=$ $\displaystyle\frac{\tan^{6}x}{6}+\frac{\tan^{8}x}{8}+C.$
###### Example 1.8.
 $\displaystyle\int\tan^{3}x\sec^{4}xdx$ $\displaystyle=$ $\displaystyle\int\tan x\tan^{2}x\sec^{4}xdx=\int\tan x(\sec^{2}x-1)\sec^{4}xdx$ $\displaystyle=$ $\displaystyle\int\tan x\sec x\sec^{5}xdx-\int\tan x\sec x\sec^{3}xdx$ $\displaystyle=$ $\displaystyle\frac{\sec^{6}x}{6}-\frac{\sec^{4}x}{4}+C.$

## 2 Trigonometric Substitutions

One can easily deduce that $\int_{0}^{1}\sqrt{1-x^{2}}dx$ has value $\frac{\pi}{4}$. Why? Simply because the graph of the function  $y=\sqrt{1-x^{2}}$ is half a circumference of radius $r=1$ (because if you square both sides of $y=\sqrt{1-x^{2}}$ you obtain $x^{2}+y^{2}=1$ which is the equation of a circle or radius $r=1$). Therefore, the area under the graph is a quarter of the area of a circle.

How does one compute $\int_{0}^{1}\sqrt{1-x^{2}}dx$ without using the geometry of the problem? This is the prototype of integral where a trigonometric substitution will work very nicely. Notice that neither substitution nor integration by parts will work appropriately.

###### Example 2.1.

Suppose we want to solve $\int_{0}^{1}\sqrt{1-x^{2}}dx$ with analytic methods. We will use a substitution $x=\sin\theta$ (so $\theta$ will be our new variable of integration), because, as we know from Equation (1), $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}\theta}=\cos\theta$, thus getting rid of the pesky square root. Notice that $dx=\cos\theta d\theta$. We also need to find the new limits of integration with respect to the new variable of integration, namely $\theta$. When $x=0=\sin\theta$ we must have $\theta=0$. Similarly, when $x=1=\sin\theta$ one has $\theta=\pi/2$. We are now ready to integrate:

 $\displaystyle\int_{0}^{1}\sqrt{1-x^{2}}dx$ $\displaystyle=$ $\displaystyle\int_{0}^{\pi/2}(\cos\theta)\cos\theta d\theta=\int_{0}^{\pi/2}% \cos^{2}\theta d\theta$ $\displaystyle=$ $\displaystyle\int_{0}^{\pi/2}\frac{1+\cos(2\theta)}{2}d\theta=\frac{1}{2}\left% (\theta+\frac{\sin(2\theta)}{2}\right)_{0}^{\pi/2}=\pi/4.$

Notice that we made use of Equation (4) in the second line.

###### Example 2.2.

Similarly, one can solve $\int_{0}^{r}\sqrt{r^{2}-x^{2}}dx$ by using a substitution $x=r\sin\theta$. Indeed, $\sqrt{r^{2}-x^{2}}=\sqrt{r^{2}-r^{2}\sin^{2}\theta}=r\cos\theta$ and $dx=r\cos\theta d\theta$. The limits of integration with respect to $\theta$ are again $\theta=0$ to $\theta=\pi/2$ (check this!). Thus:

 $\displaystyle\int_{0}^{r}\sqrt{r^{2}-x^{2}}dx$ $\displaystyle=$ $\displaystyle\int_{0}^{\pi/2}r^{2}(\cos\theta)\cos\theta d\theta=r^{2}\int_{0}% ^{\pi/2}\cos^{2}\theta d\theta$ $\displaystyle=$ $\displaystyle r^{2}\int_{0}^{\pi/2}\frac{1+\cos(2\theta)}{2}d\theta=\frac{r^{2% }}{2}\left(\theta+\frac{\sin(2\theta)}{2}\right)_{0}^{\pi/2}=r^{2}\pi/4.$

Thus, we have proved that a quarter of a circle of radius $r$ has area $r^{2}\pi/4$ which implies that the area of such a circle is $\pi r^{2}$, as usual.

The trigonometric substitutions usually work when expressions like $\sqrt{r^{2}-x^{2}}$, $\sqrt{r^{2}+x^{2}}$, $\sqrt{x^{2}-r^{2}}$ appear in the integral at hand, for some real number $r$. Here is a table of the suggested change of variables in each particular case:

If you see… try this… because…
$\sqrt{1-x^{2}}$ $x=\sin\theta$ $\sqrt{1-\sin^{2}\theta}=\cos\theta$
$\sqrt{r^{2}-x^{2}}$ $x=r\sin\theta$ $\sqrt{r^{2}-\sin^{2}\theta}=r\cos\theta$
$\sqrt{1+x^{2}}$ $x=\tan\theta$ $\sqrt{1+\tan^{2}\theta}=\sec\theta$
$\sqrt{r^{2}+x^{2}}$ $x=r\tan\theta$ $\sqrt{r^{2}+\tan^{2}\theta}=r\sec\theta$
$\sqrt{x^{2}-1}$ $x=\sec\theta$ $\sqrt{\sec^{2}\theta-1}=\tan\theta$
$\sqrt{x^{2}-r^{2}}$ $x=r\sin\theta$ $\sqrt{\sec^{2}\theta-1}=r\tan\theta$
###### Remark 2.3.

The above are “suggested” substitutions, they may not be the most ideal choice! For example, for the integral $\int 2x\sqrt{1-x^{2}}dx$, the change $u=1-x^{2}$ will work much better than $x=\sin\theta$.

###### Example 2.4.

We would like to find the value of

 $\int_{\sqrt{2}}^{2}\frac{1}{x^{3}\sqrt{x^{2}-1}}dx.$

Since neither a $u$-substitution nor integration by parts seem appropriate, we try $x=\sec\theta$, $dx=\sec\theta\tan\theta d\theta$. When $x=\sqrt{2}=\sec\theta$ one has $\theta=\pi/4$ while $x=2$ implies $\theta=\pi/3$. Hence:

 $\displaystyle\int_{\sqrt{2}}^{2}\frac{1}{x^{3}\sqrt{x^{2}-1}}dx$ $\displaystyle=$ $\displaystyle\int_{\pi/4}^{\pi/3}\frac{\sec\theta\tan\theta}{\sec^{3}\theta% \tan\theta}d\theta=\int_{\pi/4}^{\pi/3}\frac{1}{\sec^{2}\theta}d\theta=\int_{% \pi/4}^{\pi/3}\cos^{2}\theta d\theta$

and the last integral is easy to compute using Equation (4).

Title A lecture on trigonometric integrals and trigonometric substitution ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution 2013-03-22 15:38:39 2013-03-22 15:38:39 alozano (2414) alozano (2414) 4 alozano (2414) Feature msc 26A36 ALectureOnIntegrationByParts ALectureOnIntegrationBySubstitution ALectureOnThePartialFractionDecompositionMethod