Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to find the primitive function of $x^3 \sin x^2$, and I've come to a variable exchange ($t = x^2$) which led me to $\frac{1}{2} \int t \sin t dt$.

According to my text book, the primitive function is $\frac{1}{2}(-t \cos t + \sin t) + C$, but I can't see why. Isn't the derivative of that $\frac{1}{2}(t \cos t \sin t + \sin t \cos t)$?

share|cite|improve this question
Your first substitution is pretty good! Now, you just need to integrate by parts. – Mark McClure May 14 '13 at 12:39
The derivative of $(1/2)(-t \cos t +\sin t)$ is $(1/2)(-\cos t +t \sin t + \cos t)=(1/2) t \sin t$, as desired. Maybe you misused some version of the chain rule instead of using the product rule. – Tom Cooney May 14 '13 at 13:50

$\int t \sin(t) dt= \int t (-\cos(t))'dt\overset{\ast}{=}-t\cos (t) + \int -\cos(t) dt=-t\cos(t) +\sin(t) + C$, where equality $\ast$ is integration by parts.

share|cite|improve this answer

Hint: Make $u(x)=x^2$ and note that $x^3\sin x^2\mbox{d}x= u(x)\sin u(x) \frac{1}{2}u'(x) \mbox{d}x=u\sin u\, \mbox{d}u$ and use integration by parts.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.