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

#1/4#

Explanation:

First, let #u=x^2#. This implies that #du=2xdx#. When making this substitution, remember to plug the current into #u=x^2#. The bound of #0# stays #0# and #sqrtpi/2# becomes #(sqrtpi/2)^2=pi/4#.

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 #v=tan(u)# so #dv=sec^2(u)du#. The bounds become #0rarrtan(0)=0# and #pi/4rarrtan(pi/4)=1#.

#=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#