How do I integrate $\sqrt{a^2 - x^2}$?

Question

How do I integrate $$\displaystyle \int \sqrt{a^2 - x^2}dx $$

It can be solved using trig substitution, but don't know how to solve.

Thank you.

Answer 1

$\displaystyle \int \sqrt{a^2-x^2}\,dx$

Let $x=a\sin\theta \implies dx=a\cos\theta\,d\theta$

$\displaystyle \int \sqrt{a^2-a^2\sin^2\theta}\cdot a\cos\theta\,d\theta$

$=\displaystyle \int \sqrt{a^2(1-\sin^2\theta)}\cdot a\cos\theta\,d\theta$

$=\displaystyle \int a^2\cos^2\theta\,d\theta$

$=\displaystyle \int \dfrac{1}{2}a^2(1+\cos 2\theta)\,d\theta$

$=\dfrac{a^2}{2}\theta+\dfrac{a^2}{4}\sin 2\theta+C$

$=\dfrac{a^2}{2}\theta+\dfrac{a^2}{2}\sin\theta\cos\theta+C$

$\sin\theta=\dfrac{x}{a}$

$\cos\theta=\dfrac{\sqrt{a^2-x^2}}{a}$

$=\dfrac{a^2}{2}\arcsin\left(\dfrac{x}{a}\right)+\dfrac{a^2}{2}\cdot\dfrac{x}{a}\cdot\dfrac{\sqrt{a^2-x^2}}{a}+C$

$=\boxed{\dfrac{a^2}{2}\arcsin\left(\dfrac{x}{a}\right)+\dfrac{1}{2}x\sqrt{a^2-x^2}+C}$

Answer 2

Let $x=a\sin t$ with $-\frac\pi2\le t\le\frac\pi2\implies dx=a\cos t\ dt$

$t=\arcsin\dfrac xa$

$\cos t=\dfrac{\sqrt{a^2-x^2}}{|a|}$

$$\int\sqrt{a^2-x^2}dx=\int|a|a\cos^2tdt=\dfrac{a|a|}2\int(1+\cos2t)dt=\dfrac{a|a|}2\left(t+\dfrac{\sin2t}2\right)+K$$

Answer 3

We can also use integration by parts $$\left \{ \begin{matrix} u=\sqrt{a^2-x^2} \\dv=dx\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} du=\dfrac{-x}{\sqrt{a^2-x^2}}dx \\v=x.\end{matrix}\right.$$ $$I=\displaystyle\int\sqrt{a^2-x^2}dx=x\sqrt{a^2-x^2}+\displaystyle\int\dfrac{x^2}{\sqrt{a^2-x^2}}dx.\quad (1)$$ $$\begin{aligned} &\displaystyle\int\dfrac{x^2}{\sqrt{a^2-x^2}}dx=-\displaystyle\int\dfrac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx=-\displaystyle\int\sqrt{a^2-x^2}dx\\ &+a^2\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}dx=-I+a^2\operatorname{arcsin}\dfrac{x}{a}.\quad (2) \end{aligned}$$ From $(1)$ and $(2),$ we get: $$I=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\operatorname{arcsin}\dfrac{x}{a}+C.$$