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: 
$$\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,\,dx$$

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:
$$\int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta = \frac{1}{2}\ln(3 + 2\sqrt{2}) - \frac{\sqrt{2}}{2}$$
