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

1 Answer
Nov 9, 2016

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