How to rearrange the following trigonometric integral

Question

How do you rearrange the following;

$\frac{1}{R_0}\int_{0}^{\pi/2}\frac{1-sin^3(\alpha)}{cos^2(\alpha)} d\alpha$

using t=$\tan\frac{\alpha}{2}$

to obtain

$\frac{1}{R_0}\int_{0}^{1}[\frac{2}{(1+t)^2}+4\frac{t}{(1+t^2)^2}]dt$

I do not understand how this rearrangement was done

The final answer to the integral is $\frac{2}{R_0}$

Many thanks

Answer 1

The substitution you are talking about is known as the Weierstrass substitution, info on the link. It is a common method that you can derive using half angle formulae etc.

Answer 2

You may integrate directly $$\int_{0}^{\pi/2}\frac{1-\sin^3a}{\cos^2a} da = \int_{0}^{\pi/2}\frac{(1-\sin a)( 1+\sin a+\sin^2a)}{1-\sin^2a} da\\ =\int_{0}^{\pi/2}\left( \frac1{1+\sin a}+\sin a\right) da = \int_{0}^{\pi/2}\left( \frac1{1+\cos a}+\cos a\right) da\\ = \int_{0}^{\pi/2}\left( \frac12\sec^2\frac a2+\cos a\right) da = \left( \tan \frac a2+ \sin a \right)_0^{\frac\pi2}=2 $$