How do you evaluate int cosx/(1+sin^2x) from [0, pi/2]?
1 Answer
Nov 9, 2016
Explanation:
First examining without the bounds:
I=intcosx/(1+sin^2x)dx
Let
I=intsec^2theta/(1+tan^2theta)d theta=intd theta=theta+C
Since
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