# Integral by parts help

Hey Gn, I'm trying to do this exercise but I can't understand what I have to do, I suppose it is  Sin^3x^2 the derived and x the integral but I don't understand how the derivative should be in this case.

$\int x\sin^3x^2 dx$

its gonna be a big help if you guys help me, thanks for you time.

I would attempt this integral using substitution rather than IBP. The clue is that the sine has an inner function, namely the $x^2$.

First, let $u=x^2\implies du=2xdx$. Then

$\int x\sin^3(x^2)dx = \frac{1}{2}\int \sin^3(u)du = \frac{1}{2}\int \sin(u)\sin^2(u)du$

$= \frac{1}{2}\int \sin(u)\sin^2(u)du = \frac{1}{2}\int \sin(u)(1-\cos^2(u))du\;[*] = \frac{1}{2}\int \sin(u)-\sin(u)\cos^2(u)du$

$= -\frac{1}{2}\cos(u)+\frac{1}{6}\cos^3(u) = \frac{1}{6}\cos^3(x^2)-\frac{1}{2}\cos(x^2)+c$.

$[*]\;$ Using the identity $\sin^2(x)+\cos^2(x)\equiv1$

I hope this has helped you!

• Yes, thank you very much, my teacher told me that it is done by IBP but there is no practical solution, tomorrow I will talk to him if it is possible to do it that way. Again thank you very much.

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