How do you evaluate the definite integral int sec^2x/(1+tan^2x) from [0, pi/4]?
2 Answers
Aug 19, 2016
Explanation:
Notice that from the second Pythagorean identity that
This means the fraction is equal to 1 and this leaves us the rather simple integral of
Aug 20, 2016
Explanation:
Interestingly enough, we can also note that this fits the form of the arctangent integral, namely:
int1/(1+u^2)du=arctan(u)
Here, if
intsec^2x/(1+tan^2x)dx=int1/(1+u^2)du=arctan(u)=arctan(tanx)=x
Adding the bounds:
int_0^(pi/4)sec^2x/(1+tan^2x)dx=[x]_0^(pi/4)=pi/4-0=pi/4