17.5 Trigonometric Substitution

Integrals of the form $${\int F(\sqrt{a^2+x^2})dx}$$ and $${\int F(\sqrt{a^2-x^2})dx}$$ often arise in applications. There is a special trick for dealing with such integrals. Since

$$x=a\tan (\arctan ({x\over a})) \mbox{ for all } x\in\mbox{{\bf R}},$$

we can write

$$\int F(\sqrt{a^2+x^2})dx = \int F(\sqrt{a^2+\Big(a\tan(\arctan ({x\over a}))\Big)^2})dx.$$

If we now make the substitution

$$u=\arctan( {x\over a}) \mbox{ or } x=a\tan (u), \hspace{2em} \Big(u \in (-{\pi\over 2},{\pi\over 2})\Big)$$

then we find $dx=a\sec^2(u)du$, or

$$\int F(\sqrt{a^2+x^2})dx=\int F(\sqrt{a^2+\Big(a\tan (u)\Big)^2})a\sec^2 u\;du.$$

Now

$$a^2+\Big(a\tan (u)\Big)^2=a^2\Big(1+\tan^2(u)\Big)=a^2\sec^2(u)$$

so

$$\sqrt{a^2+\Big( a\tan (u)\Big)^2}=a\sec (u).$$

(Since $${u\in \Big( -{\pi\over 2},{\pi\over 2}\Big)}$$ we have $\sec (u)>0$ and the square root is positive.) Thus

$$\int F(\sqrt{a^2+x^2})dx=a\int F\Big( a\sec (u)\Big)\cdot\sec^2(u)du.$$

Often this last antiderivative can be found. If

$$a\int F\Big( a\sec (u)\Big)\cdot\sec^2(u)du = H(u),$$

then by the ritual (17.37)

$$\int F(\sqrt{a^2+x^2})dx=a\int F\Big( a\sec (u)\Big)\cdot\sec^2(u)du = H(u) = H(\arctan({x\over a})).$$

The ritual to apply when using this method for finding $${\int F(\sqrt{a^2+x^2})dx}$$ is:

Let $x=a\tan(u)$. Then $dx=a\sec^2(u)du$, and

$$\sqrt{a^2+x^2}=\sqrt{a^2+a^2\tan^2(u)}=\sqrt{a^2\sec^2(u)}=a\sec (u),$$

so

$$\int F(\sqrt{a^2+x^2})dx=a\int F\Big(a\sec (u)\Big)\sec^2(u)du = H(u) = H(\arctan({x\over a})).$$

There is a similar ritual for integrals of the form $${\int F(\sqrt{a^2-x^2})dx}$$ (Here we will just describe the ritual).

Let $x=a\sin (u)$. Then $dx=a\cos (u)du$ and

$$\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin (u)}=\sqrt{a^2\cos^2(u)}=a\cos (u)$$ (17.43)

so

$$\int F(\sqrt{a^2-x^2})dx=a\int F\Big( a\cos (u)\Big)\cdot \cos (u)du = H(u) = H(\arcsin({x\over a})).$$

Observe that in equation (17.43) we are assuming that $u=\arcsin \displaystyle {({x\over a})}$, so $${u\in ( -{\pi\over 2},{\pi\over 2})}$$, so $\cos (u)\geq 0$, and the sign of the square root is correct.

17.44   Example. Find $\int\sqrt{4+x^2}dx$.

Let $x=2\tan \theta$. Then $dx=2\sec^2\theta\;d\theta$, and

$$\sqrt{4+x^2}=\sqrt{4(1+\tan^2\theta)}=2\sqrt{\sec^2(\theta)}=2\sec(\theta).$$ (17.45)

Thus

$$\int\sqrt{4+x^2}\;dx=2^2\int\sec\theta\cdot\sec^2\theta\;d\theta= 4\int\sec^3(\theta)\;d\theta.$$

To find $\int\sec^3(\theta)d\theta$, I will integrate by parts. Let

$$\begin{array}{ll} f(\theta)=\sec(\theta), & g'(\theta) = \sec... ...theta)\tan(\theta), & g(\theta) = \tan (\theta).\\ \end{array}$$

Hence,

$$\begin{eqnarray*} \int\sec^3(\theta)d\theta &=& \int f(\theta)g'(\theta)d\theta ... ...an(\theta)-\int\sec^3(\theta)d\theta+\int\sec(\theta)d\theta.\\ \end{eqnarray*}$$

Hence

$$\begin{eqnarray*} 2\int\sec^3(\theta)d\theta &=& \sec(\theta)\tan(\theta)+\int\s... ...\theta)\tan(\theta)+\ln(\vert\sec(\theta)+\tan(\theta)\vert);\\ \end{eqnarray*}$$

i.e.,

$$\int\sec^3(\theta)d\theta={1\over 2}\Bigg(\sec(\theta)\tan(\theta)+\ln\Big(\vert\sec(\theta) +\tan(\theta)\vert\Big)\Bigg).$$ (17.46)

Hence

$$\begin{eqnarray*} \int\sqrt{4+x^2}dx &=& 4\int\sec^3(\theta)d\theta \\ &=& 2\Bi... ...(\theta)+\ln\Big(\vert\sec(\theta)+\tan(\theta)\vert\Big)\Bigg). \end{eqnarray*}$$

By (17.45) we have $\tan(\theta)=\displaystyle { {x\over 2}}$ and $${\sec(\theta) ={1\over 2}\sqrt{4+x^2}}$$. Thus

$$\begin{eqnarray*} \int\sqrt{4+x^2}dx &=& 2\Bigg( {1\over 2} \sqrt{4+x^2}\cdot{x\... ... 2}+2\ln \Big(\Big\vert {{\sqrt{4+x^2}+x}\over 2}\Big\vert\Big). \end{eqnarray*}$$

17.47   Example. In the process of working out the last example we found $\int\sec^3(\theta)d\theta$ using integration by parts. Here is a different tricky way of finding the same integral [32].

$$\begin{eqnarray*} \int \sec^3(\theta) d\theta &=& {1\over 2} \int (\sec^3(\theta... ...rt\sec(\theta) + \tan(\theta)\vert) + \sec(\theta)\tan(\theta)). \end{eqnarray*}$$

17.48   Example. Find $${ \int {1\over {\sqrt{a^2-x^2}}}dx}$$.

Let $x=a\sin(\theta)$. Then $dx=a\cos (\theta)d\theta$ and

$$\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}=a\sqrt{\cos^2\theta}=a\cos\theta.$$

Thus

$$\begin{eqnarray*} \int {1\over {\sqrt{a^2-x^2}}}dx &=& \int {{a\cos(\theta)}\ove... ...eta)}}d\theta=\int 1\;d\theta\\ &=&\theta =\arcsin({x\over a}). \end{eqnarray*}$$

17.49   Exercise. Find the following antiderivatives:

a)

$${\int\sqrt{a^2-x^2}\;dx}$$

b)

$${\int \displaystyle { {1\over {\sqrt{a^2+x^2}}}}\;dx}$$

c)

$${\int \displaystyle { {x\over {\sqrt{a^2-x^2}}}\;dx}}$$

d)

$${\int x\sqrt{a^2+x^2}\; dx}$$ A

17.50   Example (Area of a circular sector) Let $a$ be a positive number, and let $\theta_0$ be a number in $[0,{\pi\over 2})$. Let ${\bf o} = (0,0)$, and let ${\bf p} = (a\cos(\theta_0), a\sin(\theta_0)$. Let $T(a,\theta_0)$ be the circular sector bounded by the positive $x$-axis, the segment $[{\bf op]}$, and the circle $\{x^2 + y^2 = a^2\}$.

The equation for $[{\bf op}]$ is

$$y = {a\sin(\theta_0) \over a \cos(\theta_0)} x = x \tan(\theta_0),$$

and the equation for the upper semicircle is

$$y = \sqrt{a^2 - x^2}.$$

Hence

$$\mbox{\rm area}(T(a,\theta_0)) = A_0^a(f),$$

where

$$f(x) = \cases{ x\tan(\theta_0) & if $0 \leq x \leq a\cos(\th... ...\cr \sqrt{a^2 - x^2} & if $ a\cos(\theta_0) \leq x \leq a.$ }$$

i.e.

$$\mbox{\rm area}(T(a,\theta_0)) = \int_0^{a \cos(\theta_0)} x\... ...\theta_0) dx + \int_{a\cos(\theta_0)}^a \sqrt{a^2 - x^2}\; dx.$$

In exercise 17.49A.a you showed that

$$\int \sqrt{a^2 - x^2} = {1\over 2} a^2 \arcsin\left({x\over a}\right) + {1\over 2} x\sqrt{a^2 - x^2},$$

so

$$\begin{eqnarray*} \mbox{\rm area}(T(a,\theta_0)) &=& \tan(\theta_0) \left. {x^2 ... ...ver 2}\sin(\theta_0)\cos(\theta_0)\\ &=& {1\over 2}a^2\theta_0. \end{eqnarray*}$$

By using symmetry arguments, you can show that this formula actually holds for $0\leq \theta_0 \leq {2\pi}$.