Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$ Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$
I believe my work is correct until I try to change the bounds according to theta, $\theta$. 
How do I do that? Also, please comment freely on my work, that is, is my method correct and cogent. 

 A: Notice, you let  $$x=\frac{3}{5}\sin \theta\implies \theta=\sin^{-1}\left(\frac{5x}{3}\right)$$ then
upper limit at $x=0.6$ : $\theta=\sin^{-1}\left(\frac{5\times 0.6}{3}\right)=\sin^{-1}\left(1\right)=\frac{\pi}{2}$
lower limit at $x=0$ : $\theta=\sin^{-1}\left(\frac{5\times 0}{3}\right)=\sin^{-1}\left(0\right)=0$
after substitution, we get
$$\int_{0}^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\ dx=\int_{0}^{\pi/2}\frac{\left(\frac{3}{5}\sin\theta\right)^2}{\sqrt{9-9\sin^2\theta}}\frac{3}{5}\cos\theta\ d\theta$$
$$=\frac{9}{25}\cdot \frac{3}{5}\cdot \frac{1}{3}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{\sqrt{\cos^2\theta}}$$
$$=\frac{9}{125}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{|\cos\theta|}$$
we know $|\cos\theta|=\cos \theta\ \  \forall \ \ 0\le \theta \le\pi/2$
$$=\frac{9}{125}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{\cos\theta}$$
$$=\frac{9}{125}\int_{0}^{\pi/2}\sin^2\theta\ d\theta$$
A: When changing variables you should also change the integration limits! :)
$$\begin{array}{l}
\begin{array}{*{20}{l}}
{x = \frac{3}{5}\sin \theta }& \to &{\theta  = \arcsin \frac{5}{3}x}\\
{x = 0}& \to &{\theta  = \arcsin 0 = 0}\\
{x = 0.6}& \to &{\theta  = \arcsin 1 = \frac{\pi }{2}}
\end{array}\\
\end{array}$$
In fact, as $x$ varies from $0$ to $0.6$, $\theta$ varies from $0$ to $\pi \over 2$. This why that you were also allowed to write $\sqrt {{{\cos }^2}\theta }  = \left| {\cos \theta } \right| = \cos \theta $.
