# Integration techniques for $\int x^3\sin x^2\,dx$

I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

Example problem is \begin{align} \int x^3\sin x^2\,dx \end{align} I tried using a method where I set something to $u$ and $dv$ and go from there, but I don't end up anywhere with this problem. I know you can use substitution method and then integrate by parts, but I'm not sure which part of the integral I should begin substituting.

• To be clear, is this $(\sin x)^2$ or $\sin(x^2)$? The latter is discussed nicely in A-R's answer below. – Semiclassical Oct 6 '14 at 13:44

Hint:

Setting $t=x^2\,\Rightarrow\, dt=2x\,dx$, then we have \begin{align} \int x^3\sin x^2\,dx=\frac{1}{2}\int t\,\sin t\,dt \end{align} Now apply integration by part by taking $u=t$ and $dv=\sin t\,dt$.

• Hey. Thanks for the explanation. I can do the substitution by parts fine, but I'm not sure where the 1/2 is coming from. Once I have dt = 2xdx, should I solve for dx there? – user3032755 Oct 6 '14 at 13:59
• @user3032755 Because $dt=2x\,dx$ then $x\,dx=\dfrac{dt}{2}$. Hence$$\int x^3\sin x^2\,dx=\int x^2\sin (x^2)\,x\,dx$$ – Anastasiya-Romanova 秀 Oct 6 '14 at 14:12
• Thanks. Splitting the x^3 up caught me. – user3032755 Oct 6 '14 at 14:26
• @user3032755 I don't need your thankful. I need your upvote and accepting my answer if you think this is useful to you. That'll be enough payment for me. BTW, you should also accept the best answer for those who answer each your problems. You've never accepted their answers yet – Anastasiya-Romanova 秀 Oct 6 '14 at 14:36

You can also solve it, by applying integration by parts (IP) initially, then substituion.

The integral can be written as $\int x^2 \cdot x\sin(x^2)dx$, and the first IP accomplished by taking $u= x^2$ and $dv = x\sin(x^2)dx$. The second term can be integrated with the simple sub $u = x^2$.

That will leave an integral of the form $\int x \cos(x^2)dx$ (omitting constant multipliers), which can be integrated with the same sub as above.