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

Question

6787 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T15:23:31

#pi/4#

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#

Answer 2 · 2016-08-20T22:30:10

#pi/4#

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#