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

Question

1651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-27T16:13:11

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

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#