Trigonometric substitution (x = a sin θ)

With $x=a\sin\theta$ $\left(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\right)$, $a^2-x^2=a^2\cos^2\theta$, and $\cos\theta\ge 0$ on this range, so $\sqrt{a^2-x^2}=a\cos\theta$. Rewrite $dx=a\cos\theta\,d\theta$ at the same time.

As $x$ goes $0\to 1$, $\theta$ goes $0\to\frac{\pi}{2}$. When the denominator contains $a^2+x^2$, set $x=a\tan\theta$: then $a^2+x^2=\frac{a^2}{\cos^2\theta}$ and $dx=\frac{a}{\cos^2\theta}\,d\theta$, so the $\cos^2\theta$ cancels.

Examples

1. 01Which substitution removes the square root?$\displaystyle \int \sqrt{16-x^{2}}\,dx$AnswerHide

   Answer$\displaystyle x=4\sin\theta$

   Steps

   1. We want $\sin^{2}\theta+\cos^{2}\theta=1$ to apply. Setting $x=4\sin\theta$ makes the inside
   2. $\displaystyle 16-(4\sin\theta)^{2}$$\displaystyle = 16\cos^{2}\theta$
   3. That is a square, so the root comes off. With $\tan$ or a bare $\theta$ it never becomes a square
2. 02Which substitution makes the denominator a constant multiple of $\frac{1}{\cos^{2}\theta}$?$\displaystyle \int \frac{dx}{4+x^{2}}$AnswerHide

   Answer$\displaystyle x=2\tan\theta$

   Steps

   1. Here we use $1+\tan^{2}\theta=\frac{1}{\cos^{2}\theta}$. Setting $x=2\tan\theta$ makes the denominator
   2. $\displaystyle 4+(2\tan\theta)^{2}$$\displaystyle = \frac{4}{\cos^{2}\theta}$
   3. With $\sin$ you would get $1-\sin^{2}\theta$, which does not tidy the denominator
3. 03What goes in □?$\displaystyle x=3\sin\theta \;\Rightarrow\; dx = \square\,d\theta$AnswerHide

   Answer$\displaystyle 3\cos\theta$

   Steps

   1. Differentiate $x$ with respect to $\theta$
   2. $\displaystyle x$$\displaystyle = 3\sin\theta$$\displaystyle \frac{dx}{d\theta}$$\displaystyle = 3\cos\theta$
   3. $\sin$ differentiates to $\cos$; the minus sign appears only when you differentiate $\cos$
4. 04As $x$ goes $0 \to 1$, how does $\theta$ go?$\displaystyle x=\sin\theta \; \left(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\right)$AnswerHide

   Answer$\displaystyle 0 \to \frac{\pi}{2}$
5. 05What do you get after substituting?$\displaystyle \begin{gathered} \displaystyle x=2\sin\theta \; \left(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\right), \\[6pt] \displaystyle \sqrt{4-x^{2}} \end{gathered}$AnswerHide

   Answer$\displaystyle 2\cos\theta$
6. 06What goes in □?$\displaystyle x=5\sin\theta \;\Rightarrow\; dx = \square\,d\theta$AnswerHide

   Answer$\displaystyle 5\cos\theta$

   Steps

   1. Differentiate $x$ with respect to $\theta$
   2. $\displaystyle x$$\displaystyle = 5\sin\theta$$\displaystyle \frac{dx}{d\theta}$$\displaystyle = 5\cos\theta$
   3. $\sin$ differentiates to $\cos$; the minus sign appears only when you differentiate $\cos$

FAQ

Q1Why restrict $\theta$ to $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$?AnswerHide

A

On this range $\cos\theta\ge 0$, so $\sqrt{a^2\cos^2\theta}=a\cos\theta$ with no sign issue. Also, $x$ and $\theta$ correspond one to one, so the new limits are determined uniquely.

Q2When do I use $\sin$ and when $\tan$?AnswerHide

A

Use $x=a\sin\theta$ for $\sqrt{a^2-x^2}$ and $x=a\tan\theta$ for $a^2+x^2$. It depends on whether you need $1-\sin^2\theta=\cos^2\theta$ or $1+\tan^2\theta=\frac{1}{\cos^2\theta}$.