Question

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

Answer 1

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

Explanation:

`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.

• `x=pi/2=>cos(x)=cos(pi/2)=0=tan(theta)=>theta=0`
• `x=pi=>cos(x)=cos(pi)=-1=tan(theta)=>theta=-pi/4`

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`