# Definite integral

п/2
dx/(2+sinx)2
0

## 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

## Answers and Replies

hunt_mat
Homework Helper
I think if you write the integrals as:

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

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