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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 $\frac{1}{2}(t \cos t \sin t + \sin t \cos t)$?

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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

2 Answers 2

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.

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

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