Solve Definite Integral: п/2∫dx/(2+sinx)2=0

[s883]

Homework Statement

п/2
∫dx/(2+sinx)²
0

Homework Equations

The Attempt at a Solution

п/2
∫dx/(2 + sinx)² =
0
п/2
∫dx/(4 + 4sinx + (sinx)²)
0
substitude t=tg x/2 => x=2arctgx
dx=2/(1+t²) dt
sinx=2t/(1+t²)
t=tgx/2 =>
1
∫(1/(4 + 4*(2t/(1+t²)) +(2t/(1+t²))² )) * (2/(1+t²))dt =
0

1
∫( ((1+t²)²)/(2* (2(1+t²)²+4t(1+t²) +4t²) ) ) * (2/(1+t²))dt [
0

1
=∫(1+t²)/(2+4t²+2t⁴+4t+4t³+4t²) dt
0
= 1/2 ∫(1+t²)/(t⁴+2t³+4t²+2t+1) dt
And here it is divided into two integrals, but i can't come to an answer

 

[hunt_mat]

I think if you write the integrals as:

[tex]
\frac{1}{4}\int_{0}^{1}\frac{1+t^{2}}{\left[\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]^{2}}
[/tex]

Then it might give you more of an idea what to do.