Question

How do you evaluate the definite integral `int sin^2xcosx` from `[pi/4, pi/2]`?

Answer 1

`(4-sqrt2)/12`

Explanation:

`int_(pi/4)^(pi/2)sin^2xcosxdx`

We will use the substitution `u=sinx`. Note that this implies that `u=cosxdx`, both of which we have already in our integral.

Before substituting, recall to plug the current bounds into the function of `u`. The bound of `pi/4` becomes `sin(pi/4)=sqrt2/2`. The bound of `pi/2` becomes `sin(pi/2)=1`. Thus:

`=int_(sqrt2/2)^1u^2du`

Using the rule where `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=-1`:

`=[u^3/3]_(sqrt2/2)^1=1/3(1^3-(sqrt2/2)^3)=1/3(1-(sqrt2)/4)=(4-sqrt2)/12`