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 \(\PageIndex{1}\)

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

Solution

Let \(x=\sin u\), so \(dx=\cos u\,du\). Then \[\int\sqrt{1-x^2}\,dx=\int\sqrt{1-\sin^2 u}\cos u\,du=\int\sqrt{\cos^2 u}\cos u\,du.\nonumber\]

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 \(\cos u\ge 0\). Then we continue: \[\eqalign{\int\sqrt{\cos^2 u}\cos u\,du&=\int\cos^2 u\,du=\int{1+\cos 2u\over 2}\,du = {u\over 2}+{\sin 2u\over 4}+C\cr &={\arcsin x\over 2}+{\sin(2\arcsin x)\over 4}+C.\cr}\nonumber\]

This is a perfectly good answer, though the term \(\sin(2\arcsin x)\) is not very convenient. But it is possible to simplify it. Using the identity \(\sin 2x=2\sin x\cos x\), we can write \[\sin 2u=2\sin u\cos u=2\sin(\arcsin x)\sqrt{1-\sin^2 u}=2x\sqrt{1-\sin^2(\arcsin x)}=2x\sqrt{1-x^2}.\nonumber\]

Then the full antiderivative is \[\dfrac{\arcsin x}{2}+\dfrac{2x\sqrt{1-x^2}}{4}=\dfrac{\arcsin x}{2}+\dfrac{x\sqrt{1-x^2}}{2}+C.\nonumber\]

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 \(\sin^2 x+\cos^2 x=1\) in one of three forms:

\[\cos^2 x=1-\sin^2x,\qquad\sec^2 x=1+\tan^2 x,\quad\text{or}\quad\tan^2 x=\sec^2 x-1.\nonumber\]

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 \(1\).

Example \(\PageIndex{2}\)

Evaluate \(\displaystyle\int\sqrt{4-9x^2}\,dx.\)

Solution

We start by rewriting this so that it looks more like the previous example: \[\int\sqrt{4-9x^2}\,dx=\int\sqrt{4(1-(3x/2)^2)}\,dx=2\int\sqrt{1-(3x/2)^2}\,dx.\nonumber\]
Now let \(3x/2=\sin u\), so \((3/2)\,dx=\cos u\,du\) or \(dx=(2/3)\cos u\,du\). Then

\[\eqalign{2\int\sqrt{1-(3x/2)^2}\,dx&=2\int\sqrt{1-\sin^2u}\cdot{2\cos u\over 3}\,du ={4\over3}\int\cos^2 u\,du\cr &={4u\over 6}+{4\sin 2u\over12}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin u \cos u\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin(\arcsin(3x/2))\cos(\arcsin(3x/2))\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2(3x/2)\sqrt{1-(3x/2)^2}\over3}+C\cr &={2\arcsin(3x/2)\over3}+{x\sqrt{4-9x^2}\over2}+C,\cr }\nonumber\]

using some of the work from Example \(\PageIndex{1}\).

Example \(\PageIndex{3}\)

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

Solution

Let \(x=\tan u\), \(dx=\sec^2 u\,du\), so \[\int\sqrt{1+x^2}\,dx=\int\sqrt{1+\tan^2 u}\sec^2 u\,du=\int\sqrt{\sec^2 u}\sec^2 u\,du.\nonumber\]

Since \(u=\arctan(x)\), \(-\pi/2\le u\le\pi/2\) and \(\sec u\ge 0\), so \(\sqrt{\sec^2 u}=\sec u\). Then \[\int\sqrt{\sec^2 u}\sec^2 u\,du=\int\sec^3 u\,du.\nonumber\] In problems of this type, two integrals come up frequently: \(\displaystyle\int\sec^3 u\,du\) and \(\displaystyle\int\sec u\,du\). Both have relatively nice expressions but they are a bit tricky to discover.

First we do \(\displaystyle\int\sec u\,du\), which we will need to compute \(\displaystyle\int\sec^3 u\,du\): \[\eqalign{\int\sec u\,du&=\int\sec u\,\frac{\sec u +\tan u}{\sec u +\tan u}\,du\cr &=\int\frac{\sec^2 u +\sec u\tan u}{\sec u +\tan u}\,du.\cr}\nonumber\]

Now let \(w=\sec u+\tan u\), whence \(dw=\sec u\tan u+\sec^2 u\,du\), which is exactly the numerator of the function we are integrating. Thus \[\eqalign{\int\sec u\,du=\int\frac{\sec^2 u+\sec u\tan u}{\sec u+\tan u}\,du&=\int{1\over w}\,dw=\ln|w|+C\cr &=\ln|\sec u+\tan u|+C.\cr }\nonumber\]

Now for \(\displaystyle\int\sec^3 u\,du\): \[\eqalign{\sec^3 u&={\sec^3 u\over 2}+{\sec^3 u\over 2}={\sec^3 u\over 2}+{(\tan^2 u+1)\sec u\over 2}\cr &={\sec^3 u\over 2}+{\sec u\tan^2 u\over 2}+{\sec u\over 2}= {\sec^3 u+\sec u\tan^2 u\over 2}+{\sec u\over 2}.\cr }\nonumber\]

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: \[\int\sec^3 u+\sec u\tan^2 u\,du=\sec u\tan u.\nonumber\]

So putting these together we get \[\int\sec^3 u\,du={\sec u\tan u\over 2}+{\ln|\sec u+\tan u|\over 2}+C.\nonumber\] Reverting to the original variable \(x\):

\[\eqalign{\int\sqrt{1+x^2}\,dx&={\sec u\tan u\over 2}+{\ln|\sec u+\tan u|\over 2}+C\cr&={\sec(\arctan x)\tan(\arctan x)\over 2}+{\ln|\sec(\arctan x)+\tan(\arctan x)|\over2}+C\cr&={x\sqrt{1+x^2}\over 2}+{\ln|\sqrt{1+x^2}+x|\over 2}+C,\cr}\nonumber\]

using \(\tan(\arctan x)=x\) and \(\sec(\arctan x)=\sqrt{1+\tan^2(\arctan x)}=\sqrt{1+x^2}\).

Exercises \(\PageIndex{}\)

Find the antiderivatives.

Exercise \(\PageIndex{1}\)

\(\displaystyle\int\csc x\,dx\)

Answer

\(\displaystyle-\ln|\csc x+\cot x|+C=\ln\left(\tan\bigl({x\over 2}\bigr)\right)+C\)

Exercise \(\PageIndex{2}\)

\(\displaystyle\int\csc^3 x\,dx\)

Answer

\(-\dfrac{1}{2}\csc x\cot x-\dfrac{1}{2}\ln|\csc x+\cot x|+C\)

Exercise \(\PageIndex{3}\)

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

Answer

\(\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}\ln\left|x+\sqrt{x^2-1}\right|+C=\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}\text{arccosh}(x)+C\)

Exercise \(\PageIndex{4}\)

\(\displaystyle\int\sqrt{9+4x^2}\,dx\)

Answer

\(\dfrac{x}{2}\sqrt{9+4x^2}+\dfrac{9}{4}\ln\left|2x+\sqrt{9+4x^2}\right|+C\)

Exercise \(\PageIndex{5}\)

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

Answer

\(-\dfrac{1}{3}(1-x^2)^{3/2}+C\)

Exercise \(\PageIndex{6}\)

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

Answer

\(\dfrac{\arcsin x}{8}-\dfrac{\sin(4\arcsin x)}{32}+C=\dfrac{\arcsin x}{8}-\dfrac{x}{8}(1-2x^2)\sqrt{1-x^2}+C\)

Exercise \(\PageIndex{7}\)

\(\displaystyle\int{1\over\sqrt{1+x^2}}\,dx\)

Answer

\(\ln\left|x+\sqrt{1+x^2}\right|+C=\text{arcsinh}\,x+C\)

Exercise \(\PageIndex{8}\)

\(\displaystyle\int\sqrt{x^2+2x}\,dx\)

Answer

\(\dfrac{1}{2}(x+1)\sqrt{x^2+2x}-\dfrac{1}{2}\ln\left|x+1+\sqrt{x^2+2x}\right|+C=\dfrac{1}{2}(x+1)\sqrt{x^2+2x}-\dfrac{1}{2}\text{arccosh}(x+1)+C\)

Exercise \(\PageIndex{9}\)

\(\displaystyle\int{1\over x^2(1+x^2)}\,dx\)

Answer

\(-\dfrac{1}{x}-\arctan x+C\)

Exercise \(\PageIndex{10}\)

\(\displaystyle\int{x^2\over\sqrt{4-x^2}}\,dx\)

Answer

\(2\arcsin\left(\dfrac{x}{2}\right)-\dfrac{1}{2}x\sqrt{4-x^2}+C\)

Exercise \(\PageIndex{11}\)

\(\displaystyle\int{\sqrt{x}\over\sqrt{1-x}}\,dx\)

Answer

\(\arcsin(\sqrt{x})-\sqrt{x-x^2}+C=-\arcsin(\sqrt{1-x})-\sqrt{x-x^2}+C\)

Exercise \(\PageIndex{12}\)

\(\displaystyle\int{x^3\over\sqrt{4x^2-1}}\,dx\)

Answer

\(\dfrac{1+2x^2}{24}\sqrt{4x^2-1}+C\)

Exercise \(\PageIndex{13}\)

Compute \(\displaystyle\int\sqrt{x^2+1}\,dx.\quad\)(Hint: make the substitution \(x=\sinh(u)\), and then use exercise 6 in section 4.11, "Hyperbolic Functions".)

Answer

\(\dfrac{x}{2}\sqrt{x^2+1}+\dfrac{1}{2}\text{arcsinh}\,x+C=\dfrac{x}{2}\sqrt{x^2+1}+\dfrac{1}{2}\ln\left|x+\sqrt{x^2+1}\right|+C\)

Exercise \(\PageIndex{14}\)

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

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