So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a "reverse'' substitution, but it is really no different in principle than ordinary substitution.

Example 8.3.1 Evaluate $\ds\int \sqrt{1-x^2}\,dx$. Let $x=\sin u$ so $dx=\cos u\,du$. Then $$\int \sqrt{1-x^2}\,dx=\int\sqrt{1-\sin^2 u}\cos u\,du= \int\sqrt{\cos^2 u}\cos u\,du.$$ We would like to replace $\ds \sqrt{\cos^2 u}$ by $\cos u$, but this is valid only if $\cos u$ is positive, since $\ds \sqrt{\cos^2 u}$ is positive. Consider again the substitution $x=\sin u$. We could just as well think of this as $u=\arcsin x$. If we do, then by the definition of the arcsine, $-\pi/2\le u\le\pi/2$, so $\cos u\ge0$. Then we continue: \eqalign{ \int\sqrt{\cos^2 u}\cos u\,du&=\int\cos^2u\,du=\int {1+\cos 2u\over2}\,du = {u\over 2}+{\sin 2u\over4}+C\cr &={\arcsin x\over2}+{\sin(2\arcsin x)\over4}+C.\cr } This is a perfectly good answer, though the term $\sin(2\arcsin x)$ is a bit unpleasant. It is possible to simplify this. Using the identity $\sin 2x=2\sin x\cos x$, we can write $\ds \sin 2u=2\sin u\cos u=2\sin(\arcsin x)\sqrt{1-\sin^2 u}= 2x\sqrt{1-\sin^2(\arcsin x)}=2x\sqrt{1-x^2}.$ Then the full antiderivative is $${\arcsin x\over2}+{2x\sqrt{1-x^2}\over4}= {\arcsin x\over2}+{x\sqrt{1-x^2}\over2}+C.$$ $\square$

This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity $\ds \sin^2x+\cos^2x=1$ in one of three forms: $$\cos^2 x=1-\sin^2x \qquad \sec^2x=1+\tan^2x \qquad \tan^2x=\sec^2x-1.$$ If your function contains $\ds 1-x^2$, as in the example above, try $x=\sin u$; if it contains $\ds 1+x^2$ try $x=\tan u$; and if it contains $\ds x^2-1$, try $x=\sec u$. Sometimes you will need to try something a bit different to handle constants other than one.

Example 8.3.2 Evaluate $\ds\int\sqrt{4-9x^2}\,dx$. We start by rewriting this so that it looks more like the previous example: $$\int\sqrt{4-9x^2}\,dx=\int\sqrt{4(1-(3x/2)^2)}\,dx =\int 2\sqrt{1-(3x/2)^2}\,dx.$$ Now let $3x/2=\sin u$ so $(3/2)\,dx=\cos u \,du$ or $dx=(2/3)\cos u\,du$. Then \eqalign{ \int 2\sqrt{1-(3x/2)^2}\,dx&=\int 2\sqrt{1-\sin^2u}\,(2/3)\cos u\,du ={4\over3}\int \cos^2u\,du\cr &={4u\over 6}+{4\sin 2u\over12}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin u \cos u\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin(\arcsin(3x/2))\cos(\arcsin(3x/2))\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2(3x/2)\sqrt{1-(3x/2)^2}\over3}+C\cr &={2\arcsin(3x/2)\over3}+{x\sqrt{4-9x^2}\over2}+C,\cr } using some of the work from example 8.3.1. $\square$

Example 8.3.3 Evaluate $\ds\int\sqrt{1+x^2}\,dx$. Let $x=\tan u$, $\ds dx=\sec^2 u\,du$, so $$\int\sqrt{1+x^2}\,dx=\int \sqrt{1+\tan^2 u}\sec^2u\,du= \int\sqrt{\sec^2u}\sec^2u\,du.$$ Since $u=\arctan(x)$, $-\pi/2\le u\le\pi/2$ and $\sec u\ge0$, so $\ds \sqrt{\sec^2u}=\sec u$. Then $$\int\sqrt{\sec^2u}\sec^2u\,du=\int \sec^3 u \,du.$$ In problems of this type, two integrals come up frequently: $\ds \int\sec^3u\,du$ and $\ds\int\sec u\,du$. Both have relatively nice expressions but they are a bit tricky to discover.

First we do $\int\sec u\,du$, which we will need to compute $\ds \int\sec^3u\,du$: \eqalign{ \int\sec u\,du&=\int\sec u\,{\sec u +\tan u\over \sec u +\tan u}\,du\cr &=\int{\sec^2 u +\sec u\tan u\over \sec u +\tan u}\,du.\cr } Now let $w=\sec u +\tan u$, $\ds dw=\sec u \tan u + \sec^2u\,du$, exactly the numerator of the function we are integrating. Thus \eqalign{ \int\sec u\,du=\int{\sec^2 u +\sec u\tan u\over \sec u +\tan u}\,du&= \int{1\over w}\,dw=\ln |w|+C\cr &=\ln|\sec u +\tan u|+C.\cr }

Now for $\ds \int\sec^3 u\,du$: \eqalign{ \sec^3u&={\sec^3u\over2}+{\sec^3u\over2}={\sec^3u\over2}+{(\tan^2u+1)\sec u\over 2}\cr &={\sec^3u\over2}+{\sec u \tan^2 u\over2}+{\sec u\over 2}= {\sec^3u+\sec u \tan^2u\over 2}+{\sec u\over 2}.\cr } We already know how to integrate $\sec u$, so we just need the first quotient. This is "simply'' a matter of recognizing the product rule in action: $$\int \sec^3u+\sec u \tan^2u\,du=\sec u \tan u.$$

So putting these together we get $$\int\sec^3u\,du={\sec u \tan u\over2}+{\ln|\sec u +\tan u| \over2}+C,$$ and reverting to the original variable $x$: \eqalign{ \int\sqrt{1+x^2}\,dx&={\sec u \tan u\over2}+{\ln|\sec u +\tan u|\over2}+C\cr &={\sec(\arctan x) \tan(\arctan x)\over2} +{\ln|\sec(\arctan x) +\tan(\arctan x)|\over2}+C\cr &={ x\sqrt{1+x^2}\over2} +{\ln|\sqrt{1+x^2} +x|\over2}+C,\cr } using $\tan(\arctan x)=x$ and $\ds \sec(\arctan x)=\sqrt{1+\tan^2(\arctan x)}=\sqrt{1+x^2}$. $\square$

## Exercises 8.3

Find the antiderivatives.

Ex 8.3.1 $\ds\int\csc x\,dx$ (answer)

Ex 8.3.2 $\ds\int\csc^3 x\,dx$ (answer)

Ex 8.3.3 $\ds\int\sqrt{x^2-1}\,dx$ (answer)

Ex 8.3.4 $\ds\int\sqrt{9+4x^2}\,dx$ (answer)

Ex 8.3.5 $\ds\int x\sqrt{1-x^2}\,dx$ (answer)

Ex 8.3.6 $\ds\int x^2\sqrt{1-x^2}\,dx$ (answer)

Ex 8.3.7 $\ds\int{1\over\sqrt{1+x^2}}\,dx$ (answer)

Ex 8.3.8 $\ds\int\sqrt{x^2+2x}\,dx$ (answer)

Ex 8.3.9 $\ds\int{1\over x^2(1+x^2)}\,dx$ (answer)

Ex 8.3.10 $\ds\int{x^2\over\sqrt{4-x^2}}\,dx$ (answer)

Ex 8.3.11 $\ds\int{\sqrt{x}\over\sqrt{1-x}}\,dx$ (answer)

Ex 8.3.12 $\ds\int{x^3\over\sqrt{4x^2-1}}\,dx$ (answer)

Ex 8.3.13 Compute $\ds \int \sqrt{x^2 + 1}\,dx$. (Hint: make the substitution $x=\sinh(u)$ and then use exercise 6 in section 4.11.)

Ex 8.3.14 Fix $t>0$. The shaded region in the left-hand graph in figure 4.11.2 is bounded by $y=x\tanh t$, $y=0$, and $\ds x^2-y^2 =1$. Prove that twice the area of this region is $t$, as claimed in section 4.11.