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Could someone outline the step-by-step approach for the following indefinite integral?

$$\int x\sin x \ dx$$

I know the solution is $\sin(x)-x\cos(x)$, but cannot see how one would go about step-wise solving for it in a logical manner that does not include arbitrary guessing (even with intelligent guessing) of terms to use.

Thanks a bunch.

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    $\begingroup$ $\int u\, dv = uv - \int v\, du$ $\endgroup$ – abel Feb 7 '15 at 21:30
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Use Integration by parts

$$\int x \sin x dx = -x \cos x - \int \cos x dx$$

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To find (abusing notation viciously) $F(x) = \int x f(x) dx$, look at $g(x) = x\int f(x) dx $.

By the product rule, $g'(x) =xf(x)+\int f(x) dx$, so $xf(x) = g'(x)-\int f(x) dx$.

Integrating, $\int xf(x) dx =g(x)-\int \int f(x)dx $.

If $f(x) = \sin(x)$, then $\int f(x) dx = -\cos(x) $ and $g(x) = -x\cos(x) $ and $\int \int f(x) dx = -\sin(x) $ so $\int x\sin(x) dx =-x\cos(x)+\sin(x) $.

This, of course, is just integration by parts in a weird setting.

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