a way to integrate $ \int \frac {\sin^2t}{\cos^3t} dt$?

Question

is there a way to integrate $$ \int \frac {\sin^2t}{\cos^3t} dt$$ that would not be too difficult or complex?

Answer 1

HINT: Multiply numerator and denominator by $\cos t$ and let $u=\sin t$. You obtain $$ \int\frac{\sin^2t}{(1-\sin^2t)^2}\cos t\,dt $$

Answer 2

Do you know how to integrate $\sec x$? If so then $$ \int \frac{\sin^2 x}{\cos^3 x}dx = -\int \frac{\sin x}{3}\frac{d}{dx}\sec^2 x dx $$ then use by parts.

Answer 3

The integrand equals $\tan^2 \sec = (\sec^2 - 1) \sec = \sec^3 - \sec$

Here you go:

http://www.math.ubc.ca/~feldman/m121/secx.pdf http://en.wikipedia.org/wiki/Integral_of_secant_cubed