Question

How do you evaluate `int cosx/(1+sin^2x)` from `[0, pi/2]`?

Answer 1

`intcosx/(1+sin^2x)dx=pi/4`

Explanation:

First examining without the bounds:

`I=intcosx/(1+sin^2x)dx`

Let `sinx=tantheta`. This may look like a wild substitution, but the goal is to get the denominator of the fraction to `1+tan^2theta=sec^2theta`. Note that differentiating `sinx=tantheta` gives `cosxdx=sec^2thetad theta`. It's also helpful that `cosxdx` is already present in the integrand. Substituting these in:

`I=intsec^2theta/(1+tan^2theta)d theta=intd theta=theta+C`

Since `sinx=tantheta`, we see that `theta=arctan(sinx)`:

`I=arctan(sinx)+C`

Adding the bounds:

`I_B=int_0^(pi/2)cosx/(1+sin^2x)dx=[arctan(sinx)]_0^(pi/2)`

`color(white)(I_B)=arctan(sin(pi/2))-arctan(sin(0))`

`color(white)(I_B)=arctan(1)-arctan(0)`

`color(white)(I_B)=pi/4-0`

`color(white)(I_B)=pi/4`