definite intergrals

Question

i couldn't add certain signs of calculus hence a verbal mathematical illustration of the question being asked of you geniuses

the integral with limits 0 to pie/2 sin² x / (1+cosx)² dx

Answer 1

Firstly let's "desquare " the situation.

∫₀^π/2 sin²x / (1 + cos²x ) dx = ∫₀^π/2 (1-cos 2x )/(2 +cos 2x) d x = (1/2) ∫₀^π (1-cos u )/(2 +cos u) d u.

Here you need the Weirstrass substitution

Let t = tan(u/2). Then if you sketch a right triangle you will get cos(u/2) =1/√(1 +t²) and

sin(u/2) = t/√(1 +t²).

Then you can prove that cosu= ( 1 -t²)/ (1 +t² ) and sinu = ( 2t)/ (1 +t² )

Also it's easy to show that du = [ 2/( 1+t²)] dt

Then the ∫ (1-cos u )/(2 +cos u) d u = ∫{ [ 1- ( 1 -t²)/ (1 +t² ) ] / [ 2 + ( 1 -t²)/ (1 +t² ) ] } [ 2/( 1+t²)] dt =

= 4∫ [ t⁴ +3t² ] / [ (1+t²)³] dt

Then you will need the partial fraction decomposition

[ t⁴ +3t² ] / [ (1+t²)³] = 1 / (1+x² ) + 1 / (1+x²)² − 2 / (1+ t²)³

I hope you can take it from there.

I would have continued but I am going through a surgical procedure tomorrow and I need to get some rest.

In actuality the problem is slightly easier since the integrant sin² x / (1+cosx)² simplifies faster

as sin² x / (1+cosx)² = ( 1- cos²x ) / ( 1+cosx)² = ( 1-cosx) / (1+ cosx ) but the Weierstrass substitution is still valid