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: 
$$\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,\,dx$$
 A: 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.  
A: You may observe that, with $u=\sin \theta$, we have
$$
\int_0^{\large \pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta=\int_0^{\large \pi/4}\frac{\sin^2 \theta}{\cos^2 \theta}\cos \theta \:d\theta=\int_0^{ \sqrt{2}/2}\frac{u^2 }{1-u^2} \:du=\log(\sqrt{2}+1)-\frac{\sqrt{2}}2.
$$
A: Use the trick to add and remove a $\cos^2\theta$ term in the numerator of the integral:
$$\int_0^{\pi/4}\frac{\sin^2\theta + \cos^2\theta - \cos^2\theta}{\cos\theta}\ \text{d}\theta = \int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta$$
Which now you can split.
Remembering that $$\frac{1}{\cos\theta} = \text{sec}(\theta)$$
you have immediately, by remembering standard integral of the secant:
$$\int_0^{\pi/4}\text{sec}\theta\ \text{d}\theta = \left[\ln\left(\cos\frac{\theta}{2} - \sin\frac{\theta}{2}\right) + \ln\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right)\right]\bigg|_0^{\pi/4} = \frac{1}{2}\ln(3 + 2\sqrt{2})$$
The other one is trivial:
$$\int_0^{\pi/4}\cos\theta\ \text{d}\theta = \frac{\sqrt{2}}{2}$$
Thence:
$$\int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta = \frac{1}{2}\ln(3 + 2\sqrt{2}) - \frac{\sqrt{2}}{2}$$
