Integration Trig Sub

Idea of trigonometric substitution: almost the reverse of $u$-substitution, we want to add trigonometry to a problem so that the integral simplifies, using trigonometric identities. It’s best to see this with an example:

$$\int \sqrt{1-x^2}dx$$

This is the area under a…..semicircle! Since the curve is related to circles, we could try to replace $x$ with a trigonometric function, which measures the $x$-coordinate of a circle $x=\cos \theta$. Then $dx=-\sin \theta d\theta$, and so

$$\int \sqrt{1-x^2}=\int \sqrt{1-{\cos }^{2}x}(-\sin \theta )d\theta$$ This square root simplifies to $|\sin \theta |$ (using ${\cos }^2+{\sin }^2=1$), and so all together the substitution results in $-\int |\sin \theta |\sin \theta d\theta$ Which, when $\sin$ is positive (on $[0,\pi ]$, say) is just the integral $$-\int {\sin }^2\theta d\theta$$ Which is one we have already learned to do with trigonometric identities!

What substitutions are useful?

This is the overall goal:

• Choose a substitution $x=$trigonometric function

• Compute $dx$, substitute all into the integral

• Use trigonometric identities to simplify the result, so you can integrate.

The difficulty in this technique is not the substitution or integration itself (its just a form of $u$-substitution, after all!) but rather making a good choice of which substitution to make in the first place. We want the choice to lead to some nice simplification using trigonometric identities, which means we need to be thinking ahead!We derive three useful substitutions from two identities ${\sin }^2+{\cos }^2=1$ and ${\sec }^2=1+{\tan }^2$, by re-arranging and taking square roots.

• $\sqrt{1-{\sin }^2\theta }=|\cos \theta |$ means that the substitution $x=\sin \theta$ will convert expressions like $\sqrt{1-x^2}$ to $|\cos \theta |$.
• $\sqrt{1+{\tan }^2\theta }=|\sec \theta |$ means that the substitution $x=\tan \theta$ will convert expressions like $\sqrt{1+x^2}$ to $|\sec \theta |$.
• $\sqrt{{\sec }^2\theta -1}=|\tan \theta |$ means that the substitution $x=\sec \theta$ will convert expressions like $\sqrt{x^2-1}$ to $|\tan \theta |$.

Substituting $x=a\sin \theta$

When an integral contains $\sqrt{a^2-x^2}$ (where $x$ ranges over a subset of $[-a,a]$), we may wish to substitute $x=a\sin \theta$ (where $\theta$ ranges over a subset of $[-\pi /2,\pi /2]$). This converts $\sqrt{a^2-x^2}$ into $|a\cos \theta |$ and $dx=a\cos \theta d\theta$.

Examples: $$\int \frac{\sqrt{9-x^2}}{x^2}dx$$ $$x=3\sin \theta$$

Then $\cot \theta =\frac{9-x^2}{x}$ so finally $$\int \frac{\sqrt{9-x^2}}{x^2}dx=-\frac{\sqrt{9-x^2}}{x}-\arcsin \frac{x}{3}+C$$

Substituting $x=a\tan \theta$

When an integral contains $\sqrt{a^2+x^2}$ (where $x$ ranges over any subset of the real line), we may wish to substitute $x=a\tan \theta$ (where $\theta$ ranges over a subset of $[-\pi /2,\pi /2]$). This converts $\sqrt{a^2+x^2}$ into $|a\sec \theta |$ and $dx=a{\sec }^2\theta d\theta$.

$$\int \frac{1}{x^2\sqrt{x^2+4}dx$$$$x=2\tan\theta$$

Then $\csc \theta =\frac{\sqrt{x^2+4}}{x}$ so finally $$\int \frac{dx}{x^2\sqrt{x^2+4}}=-\frac{\sqrt{x^2+4}}{4x}+C$$

Substituting $x=a\sec \theta$

To do these integrals, its often useful to know the integral of secant $$\int \sec (x)dx=\ln |\sec x+\tan x|+C$$ This can be confirmed by differentiation; though we will derive it soon when we do partial fractions!

When an integral contains $\sqrt{x^2-a^2}$ (where $x$ ranges is always greater in absolute value than $a$), we may wish to substitute $x=a\sec \theta$ (where $\theta$ ranges over a subset of $[0,\pi /2]$ or $[\pi ,\frac{3}{2}\pi ]$). This converts $\sqrt{x^2-a^2}$ into $|a\tan \theta |$ and $dx=a\sec \theta \tan \theta d\theta$.

$$\int \frac{1}{\sqrt{x^2-2}}dx$$ Let $x=\sqrt{2}\sec \theta$.

Substitution for Definite Integrals

If an integral is a definite integral measuring the area under a curve, we can convert the bounds of integration instead of transforming back to the original variable, and then evaluating at the original bounds. While in ‘standard’ $u$-sub both approaches take the same amount of work, converting the bounds here feels significantly easier than going through with the whole substitution first!

Here’s an example:

$${\int }_0^3\frac{x}{\sqrt{36-x^2}}dx$$

Making the substitution $x=6\sin \theta$ on the indefinite integral results in $$\int \frac{6\sin \theta }{6\cos \theta }6\cos \theta d\theta =\int 6\sin \theta d\theta$$

Substituting the bounds to get our new definite integral, we need to figure out what $\theta$ equals when $x=0$ and $x=3$. If $x=6\sin \theta =0$ and $\theta \in [-\pi /2,\pi /2]$, then $\theta =0$. If $x=6\sin \theta =3$ then $\sin \theta =1/2$, which has the unique solution $\theta =\pi /6$ for $\theta \in [-\pi /2,\pi /2]$. Thus our new definite integral is $$6{\int }_0^{\frac{\pi }{6}}\sin \theta d\theta$$

Recognizing a ‘Hidden’ Trigonometric Substitution

$$\int \sqrt{3+2x-x^2}dx$$

Complete the square underneath the integral before proceeding! $$3+2x-x^2=4-(x-1{)}^2$$ Thus our integral is $$\int \sqrt{4-(x-1{)}^2}dx$$ Substituting $u=x-1$ will make the square root into a form we know: $$\int \sqrt{4-u^2}du$$ This we can attack with the trigonometric substitution $u=2\sin \theta$!