Integrate $x^2\sqrt{a^2-x^2}$

Question

I wanna integrate $$\int x^2 \sqrt{a^2-x^2} dx$$ but I don't know what method to apply.

I think some substitution is favorable, perhaps trigonometric, but I don't know which. Help would be appreciated.

Answer 1

Hint: Let $x = a\sin t$ then $dx = a \cos t dt$.

$$\int a^3\sin^2 t \sqrt{a^2 - a^2\sin^2 t}\cos t\ \ dt = \int a^4\sin^2 t \cos^2 t \ \ dt =\frac{1}{4}\int a^4(\sin^2 2t) \ \ dt $$

Use $\cos 2t = 1 - 2\sin^2 t$.

Answer 2

Hint The expression $\sqrt{a^2 - x^2}$ suggests using

• the trigonometric substitution $$x = a \sin \theta, \qquad dx = a \cos \theta \,d\theta ,$$ which transforms the integral to $$a^4 \int \sin^2 \theta \cos^2 \theta \,d\theta,$$ or
• the hyperbolic substitution $$x = a \tanh t, \qquad dx = a \operatorname{sech}^2 t,$$ which transforms the integral to $$a^4 \int \frac{\sinh^2 t}{\cosh^5 t} \,dt .$$

Answer 3

When you see an integral first thing that you might do if you want to do it by substitution is that check the domain of $x$ and see if any common function fits that domain. In this integral, the function is $\sqrt{a^2-x^2}$. Hence $x$ can vary between $+a$ and $-a$. So you can easily see that $a\sin\theta$ or $a\cos\theta$ fits the domain. Using $x=a\sin\theta$ and $dx=a\cos\theta d\theta$, you can change the integral as $$\begin{align}\int a^2\sin^2\theta*a\cos\theta*a\cos\theta d\theta&=\frac {a^4}4\int4\sin^2\theta\cos^2\theta d\theta\\&=\frac {a^4}4\int \sin^22\theta d\theta\\&=\frac{a^4}8\int (1-\cos4\theta) d\theta\\&=\frac {a^4}8\left|\theta-\frac {sin4\theta}4\right|\\&=\frac {a^4\sin^{-1}\frac xa}8-\frac{a^4\sin \left(4*\sin^{-1}\frac xa\right)}{32}\end{align}$$