Integral of #int x/sin^2(x^2)dx#?

1 Answer
Jun 27, 2018

#int x/sin^2(x^2)dx = -1/2cot(x^2) + c#

Explanation:

Let #u = x^2#, then #1/2u = xdx#:

#1/2int 1/sin^2(u)du#

Substitute #1/sin^2(u) = csc^2(u)#:

#1/2int csc^2(u)du#

#-1/2cot(u) + c#

Reverse the substitution:

#int x/sin^2(x^2)dx = -1/2cot(x^2) + c#