How do you integrate $\int x sin 2 x$ $int xsin2x$ by integration by parts method?

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How do you integrate $\int x sin 2 x$ $int xsin2x$ by integration by parts method?

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Answer 1 · 2017-01-16T11:46:55

$\int x sin 2 x d x = - \frac{1}{2} x cos 2 x + \frac{1}{4} sin 2 x$ $int xsin2xdx = -1/2xcos2x +1/4sin2x$

Explanation:

As:

$d (cos 2 x) = - 2 sin 2 x d x$ $d(cos2x) =-2sin2x dx$

we can integrate by parts in this way:

$\int x sin 2 x d x = - \frac{1}{2} \int x d (cos 2 x) = - \frac{1}{2} x cos 2 x + \frac{1}{2} \int cos 2 x d x = - \frac{1}{2} x cos 2 x + \frac{1}{4} sin 2 x$ $int xsin2xdx = -1/2 int x d(cos2x) = -1/2 xcos2x +1/2 int cos2x dx= -1/2xcos2x +1/4sin2x$

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How do you integrate $\int x sin 2 x$ $int xsin2x$ by integration by parts method?

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