Integration trigometry?

Question

how to integrate:

x / [sin^2 (x^2)]

Answer 1

Let u = x². Then du = 2x dx. So we get
½∫ du / sin² u = ½∫ csc² u du = -½ cot u = -½cot(x²) + C.

Answer 2

This is ******* easy!!!! take x^2 to be t and multiply numerator and denominator by -2 take the -2 in denominator out of the intergral you will get -1/2{-2xdx/sin^t
we can write dt as 2xdx therefore it will be reduced to -1/2{-dt/sint^2 write -1/sint^2 as -cosect^2 and intergral of -cosect^2 is cot t therefore you will get -1/2cotx^2+C where C is constant of intergration . (Just remember i have represented intregal sign with { )