A lecture on trigonometric integrals and trigonometric substitution

1 Trigonometric Integrals

First, we must recall a few trigonometric identities:

${\sin }^{2}x+{\cos }^{2}x$ $=$ $1$ (1)

${\sec }^{2}x$ $=$ $1+{\tan }^{2}x$ (2)

${\sin }^{2}x$ $=$ ${\displaystyle \frac{1-\cos (2x)}{2}}$ (3)

${\cos }^{2}x$ $=$ ${\displaystyle \frac{1+\cos (2x)}{2}}$ (4)

$\sin (2x)$ $=$ $2\sin x\cos x$ (5)

$\cos (2x)$ $=$ ${\cos }^{2}x-{\sin }^{2}x.$ (6)

The most usual integrals which involve trigonometric functions can be solved using the identities above.

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 $\int {\cos }^{m}x\sin xdx$ or $\int {\sin }^{m}x\cos xdx$ 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.

${\displaystyle \int {\sin }^{3}xdx}$ $=$ ${\displaystyle \int {\sin }^{2}x\sin xdx}={\displaystyle \int (1-{\cos }^{2}x)\sin xdx}$

$=$ ${\displaystyle \int \sin xdx}-{\displaystyle \int {\cos }^{2}x\sin xdx}$

$=$ $-\cos x+{\displaystyle \frac{{\cos }^{3}x}{3}}+C.$

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

Example 1.5.

${\displaystyle \int {\cos }^{3}x{\sin }^{2}xdx}$ $=$ ${\displaystyle \int {\cos }^{2}x\cos x{\sin }^{2}xdx}={\displaystyle \int (1-{\sin }^{2}x)\cos x{\sin }^{2}xdx}$

$=$ ${\displaystyle \int \cos x{\sin }^{2}xdx}-{\displaystyle \int \cos x{\sin }^{4}xdx}$

$=$ ${\displaystyle \frac{{\sin }^{3}x}{3}}-{\displaystyle \frac{{\sin }^{5}x}{5}}+C.$

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.

${\displaystyle \int {\tan }^{5}x{\sec }^{4}xdx}$ $=$ ${\displaystyle \int {\tan }^{5}x{\sec }^{2}x{\sec }^{2}xdx}={\displaystyle \int {\tan }^{5}x(1+{\tan }^{2}x){\sec }^{2}xdx}$

$=$ ${\displaystyle \int {\tan }^{5}x{\sec }^{2}xdx}+{\displaystyle \int {\tan }^{7}x{\sec }^{2}xdx}$

$=$ ${\displaystyle \frac{{\tan }^{6}x}{6}}+{\displaystyle \frac{{\tan }^{8}x}{8}}+C.$

Example 1.8.

${\displaystyle \int {\tan }^{3}x{\sec }^{4}xdx}$ $=$ ${\displaystyle \int \tan x{\tan }^{2}x{\sec }^{4}xdx}={\displaystyle \int \tan x({\sec }^{2}x-1){\sec }^{4}xdx}$

$=$ ${\displaystyle \int \tan x\sec x{\sec }^{5}xdx}-{\displaystyle \int \tan x\sec x{\sec }^{3}xdx}$

$=$ ${\displaystyle \frac{{\sec }^{6}x}{6}}-{\displaystyle \frac{{\sec }^{4}x}{4}}+C.$

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:

${\displaystyle {\int }_0^1}\sqrt{1-x^2}𝑑x$ $=$ ${\displaystyle {\int }_0^{\pi /2}}(\cos \theta )\cos \theta d\theta ={\displaystyle {\int }_0^{\pi /2}}{\cos }^2\theta d\theta$

$=$ ${\displaystyle {\int }_0^{\pi /2}}{\displaystyle \frac{1+\cos (2\theta )}{2}}𝑑\theta ={\displaystyle \frac{1}{2}}{\left(\theta +{\displaystyle \frac{\sin (2\theta )}{2}}\right)}_0^{\pi /2}=\pi /4.$

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:

${\displaystyle {\int }_0^r}\sqrt{r^2-x^2}𝑑x$ $=$ ${\displaystyle {\int }_0^{\pi /2}}{r}^2(\cos \theta )\cos \theta d\theta =r^2{\displaystyle {\int }_0^{\pi /2}}{\cos }^2\theta d\theta$

$=$ $r^2{\displaystyle {\int }_0^{\pi /2}}{\displaystyle \frac{1+\cos (2\theta )}{2}}𝑑\theta ={\displaystyle \frac{r^2}{2}}{\left(\theta +{\displaystyle \frac{\sin (2\theta )}{2}}\right)}_0^{\pi /2}=r^2\pi /4.$

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 $\mathrm{\int }\mathrm{2}\mathit{}x\mathit{}\sqrt{\mathrm{1}\mathrm{-}{x}^{\mathrm{2}}}\mathit{}𝑑x$, the change $u\mathrm{=}\mathrm{1}\mathrm{-}{x}^{\mathrm{2}}$ will work much better than $x\mathrm{=}\sin \mathit{}\theta$.

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:

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

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

Title	A lecture on trigonometric integrals and trigonometric substitution
Canonical name	ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution
Date of creation	2013-03-22 15:38:39
Last modified on	2013-03-22 15:38:39
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	4
Author	alozano (2414)
Entry type	Feature
Classification	msc 26A36
Related topic	ALectureOnIntegrationByParts
Related topic	ALectureOnIntegrationBySubstitution
Related topic	ALectureOnThePartialFractionDecompositionMethod