<s>[INST] How do I finish this trig integral $\int_0^{\pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta$? I got up to the part where it's $$\frac{9}{125}\int_0^{\large \frac{\pi}{4}}\frac{\sin^2\theta}{\cos\theta}\,\,d\theta$$
but I can't figure out how to finish it off.
By the way the original problem was: 
$$\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,\,dx$$
 [/INST] Your substitution is a bit off. Begin with
$$\begin{align*}
\int_0^{0.6}{\frac{x^2}{\sqrt{9-25x^2}}}\, dx
&= \int_0^{0.6}{\frac{x^2}{\sqrt{25\left(\frac{9}{25}-x^2\right)}}}\, dx \\
&= \frac{1}{5}\int_0^{0.6}{\frac{x^2}{\sqrt{\frac{9}{25}-x^2}}}\, dx.
\end{align*}
$$
Now let 
$$\begin{align*}
x&=\frac{3}{5}\sin \theta \\
dx&=\frac{3}{5}\cos\theta \, d \theta. \\
x^2&=\frac{9}{25}\sin^2\theta, \\
\frac{9}{25}-x^2&=\frac{9}{25}-\frac{9}{25}\sin^2\theta\\
&=\frac{9}{25}\cos^2\theta.
\end{align*}
$$
To change limits, note that when $x=0$, $\frac{3}{5}\sin \theta=0\Rightarrow \theta=0$, but when $x=0.6=\frac{3}{5}$, $\frac{3}{5}\sin \theta=\frac{3}{5}\Rightarrow \sin\theta=1 \Rightarrow \theta=\frac{\pi}{2}$.
So your integral is now
$$\frac{1}{5}\int_0^{\frac{\pi}{2}}\frac{\frac{9}{25}\sin^2\theta\frac{3}{5}\cos \theta}{\sqrt{\frac{9}{25}\cos^2\theta}}\, d \theta.$$
I'll let you pick it up from there.  
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