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 canbe solved using the identities above.

Example 1.1.

$\int \sin xdx=-\cos x+C$ and $\int \cos xdx=\sin x+C$ areimmediate 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 theintegral 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 firstas $\int {\cos }^{5}x{\sin }^{2}x\sin x=\int {\cos }^{5}x(1-{\cos }^{2}x)\sin xdx$ and then proceed as in the previous example.

One can use similar tricks to solve integrals which involveproducts 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 $r=1$ (becauseif 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 acircle.

How does one compute ${\int }_0^1\sqrt{1-x^2}𝑑x$ without using thegeometry of the problem? This is the prototype of integral where atrigonometric substitution will work very nicely. Notice thatneither substitution nor integration by parts will workappropriately.

Example 2.1.

Suppose we want to solve ${\int }_0^1\sqrt{1-x^2}𝑑x$ with analyticmethods. We will use a substitution $x=\sin \theta$ (so $\theta$will be our new variable of integration), because, as we know fromEquation (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 ofintegration 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 arenow 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 asubstitution $x=r\sin \theta$. 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 respectto $\theta$ 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 $r$ hasarea $r^2\pi /4$ which implies that the area of such a circle is$\pi r^2$, as usual.

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

Remark 2.3.

The above are “suggested” substitutions, they may not be themost 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 seemappropriate, 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).