Solving Integral of x/(sin(x^2))^2: Tips and Tricks

[Redoctober]

V = \int \frac{x}{(sin(x^2))^2}.dx

I tried to solve this by parts or by substitution but that doesn't seem to work.
Any Idea how to solve it :/

 

[HallsofIvy]

Looks to me like the substitution u= x^2 is obvious. With that,
\int \frac{x}{sin^2(x^2)}dx= \frac{1}{2}\int \frac{1}{sin^2(u)}du= \frac{1}{2}\int csc^2(x)dx
That's fairly straightforward.

 

[Abhinav R]

∫x/sin(x²)² dx
Put x²=t
∴ 2x dx=dt
x dx = dt/2
Substituting we get,
1/2∫dt/sin t²
= 1/2∫cosec²t dt
= -1/2 cot t + C
= -1/2 cot²x + C