How do you find the antiderivative of #int xsec^2(x^2)tan(x^2)dx# from #[0,sqrtpi/2]#?
1 Answer
Oct 21, 2016
Explanation:
First, let
Thus:
#int_0^(sqrtpi/2)xsec^2(x^2)tan(x^2)dx=1/2int_0^(pi/4)sec^2(u)tan(u)du#
Here notice that the derivative of tangent is present alongside the tangent function. Let
#=1/2int_0^1vdv=1/2[v^2/2]_0^1=1/2(1^2/2-0^2/2)=1/2(1/2)=1/4#