How do you evaluate $int sinx/(1+cos^2x)$ from $[pi/2, pi]$ ?

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Answer 1 · 2017-01-09T22:02:40

$int_(pi//2)^pisinx/(1+cos^2x)dx=pi/4$

$I=int_(pi//2)^pisinx/(1+cos^2x)dx$

We will make the substitution $cosx=tantheta$ . Differentiating this shows that $-sinxcolor(white).dx=sec^2thetacolor(white).d theta$ .

When making this substitution from $x$ to $theta$ , we also need to change the bounds.

So:

$I=-int_(pi//2)^pi(-sinxcolor(white).dx)/(1+cos^2x)=-int_0^(-pi//4)(sec^2thetacolor(white).d theta)/(1+tan^2theta)$

Reversing the order of the bounds with the negative sign and using the identity $tan^2theta+1=sec^2theta$ :

$I=int_(-pi//4)^0d theta=[theta]_(-pi//4)^0=0-(-pi/4)=pi/4$

How do you evaluate int sinx/(1+cos^2x)int sinx/(1+cos^2x) from [pi/2, pi][pi/2, pi]?