How do you evaluate the definite integral int sec^2x/(1+tan^2x) from [0, pi/4]?

2 Answers
Aug 19, 2016

pi/4

Explanation:

Notice that from the second Pythagorean identity that

1 + tan^2x = sec^2x

This means the fraction is equal to 1 and this leaves us the rather simple integral of

int_0^(pi/4)dx = x|_0^(pi/4) = pi/4

Aug 20, 2016

pi/4

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 u=tanx then du=sec^2xdx, then:

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