Calculate : $\int_{-\pi/2}^{\pi/2}\int_0^{asin\theta} r^2 drd\theta$

Question

I need help with the following integral :

$$\int_{-\pi/2}^{\pi/2}\int_0^{asin\theta} r^2 drd\theta$$

What I did :

$$\int_{-\pi/2}^{\pi/2}\int_0^{asin\theta} r^2 drd\theta=\int_{-\pi/2}^{\pi/2}\frac{1}{3}\int_0^{asin\theta} r^3 drd\theta=\int_{-\pi/2}^{\pi/2}(asin\theta)^3 d\theta$$

What is the integral of $(asin\theta)^3$? Any Ideas?

Thanks in advence!

Answer 1

The last integral should be $\mathrm{d} \theta$ I believe. Also, since you're integrating an odd function over a symmetric interval, the result should be zero.

Answer 2

There is a small mistake in your process. It should be:

$$\int_{-\pi/2}^{\pi/2}\int_0^{asin\theta} r^2 drd\theta=\int_{-\pi/2}^{\pi/2}\frac{1}{3}(asin\theta)^3 d\theta=0$$

Answer 3

Assuming $a$ is a constant, $\sin^3(\theta)$ can be broken down into $\sin^2(\theta)\sin(\theta)$. $\sin^2(\theta)$ can be simplified using the identity $\cos(2\theta)=1-2\sin^2(\theta)$. Then use integration by parts to solve it.