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

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SUMMARY

The discussion focuses on solving the definite integral \(\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{dx}{(2 + \sin x)^2} = 0\). The solution involves a substitution \(t = \tan \frac{x}{2}\), leading to a transformed integral that simplifies the expression. Participants explore the integral's decomposition into two parts, ultimately suggesting a reformulation to \(\frac{1}{4} \int_0^1 \frac{1 + t^2}{\left[\left(t + \frac{1}{2}\right)^2 + \frac{3}{4}\right]^2} dt\) for easier evaluation.

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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

 
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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.
 

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