Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.3: Trigonometric Substitutions

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 \(\int \sqrt{1-x^2}\,dx\).


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 \( \sqrt{\cos^2 u}\) by \(\cos u\), but this is valid only if \(\cos u \) is positive, since \( \sqrt{\cos^2 u}\) 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 \(\sin 2u=2\sin u\cos u=2\sin(\arcsin x)\sqrt{1-\sin^2 u}= 2x\sqrt{1-\sin^2(\arcsin x)}=2x\sqrt{1-x^2}.\) Then the full antiderivative is $$ {\arcsin x\over2}+{2x\sqrt{1-x^2}\over4}= {\arcsin x\over2}+{x\sqrt{1-x^2}\over2}+C. $$

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 \(\sin^2x+\cos^2x=1\) 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 \(1-x^2\), as in the example above, try \(x=\sin u\); if it contains \(1+x^2\) try \(x=\tan u\); and if it contains \( x^2-1\), 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 \(\int\sqrt{4-9x^2}\,dx\).


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.

Example 8.3.3

Evaluate \(\int\sqrt{1+x^2}\,dx\).


Let \(x=\tan u\), \(dx=\sec^2 u\,du\), 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 \(\sqrt{\sec^2u}=\sec u\). Then $$\int\sqrt{\sec^2u}\sec^2u\,du=\int \sec^3 u \,du.$$ In problems of this type, two integrals come up frequently: \( \int\sec^3u\,du\) and \(\int\sec u\,du\). 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 \(\int\sec^3u\,du$\): $$\eqalign{ \int\sec u\,du&=\int\sec u\,{\sec u +\tan u\over \sec u +\tan u}\,du\cr &=\int{\sec^2 u +\sec u\tan u\over \sec u +\tan u}\,du.\cr }$$ Now let \(w=\sec u +\tan u\), \(dw=\sec u \tan u + \sec^2u\,du\), 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 \(\int\sec^3 u\,du\): $$\eqalign{ \sec^3u&={\sec^3u\over2}+{\sec^3u\over2}={\sec^3u\over2}+{(\tan^2u+1)\sec u\over 2}\cr &={\sec^3u\over2}+{\sec u \tan^2 u\over2}+{\sec u\over 2}= {\sec^3u+\sec u \tan^2u\over 2}+{\sec u\over 2}.\cr }$$ 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 \(\sec(\arctan x)=\sqrt{1+\tan^2(\arctan x)}=\sqrt{1+x^2}\).


  • Integrated by Justin Marshall.