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Let $I=\int x \sin x d x$

Taking x as first function and sin x as second function and integrating by parts, we obtain

$\begin{aligned} I &=x \int \sin x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sin x d x\right\} d x \\ &=x(-\cos x)-\int 1 \cdot(-\cos x) d x \\ &=-x \cos x+\sin x+\mathrm{C} \end{aligned}$