# How do you integrate #int xsinxcosx# by integration by parts method?

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To integrate ( \int x \sin(x) \cos(x) ) using integration by parts:

Let ( u = x ) and ( dv = \sin(x) \cos(x) , dx ). Then ( du = dx ) and ( v = -\frac{1}{2} \sin^2(x) ).

Using the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int x \sin(x) \cos(x) , dx = -\frac{1}{2}x \sin^2(x) - \int (-\frac{1}{2} \sin^2(x)) , dx ]

Now, integrate ( -\frac{1}{2} \sin^2(x) ):

[ \int (-\frac{1}{2} \sin^2(x)) , dx = \frac{1}{2} \int \sin^2(x) , dx ]

This integral can be solved using the half-angle identity ( \sin^2(x) = \frac{1 - \cos(2x)}{2} ):

[ \frac{1}{2} \int \sin^2(x) , dx = \frac{1}{2} \int \frac{1 - \cos(2x)}{2} , dx ]

[ = \frac{1}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

Putting it all together:

[ \int x \sin(x) \cos(x) , dx = -\frac{1}{2}x \sin^2(x) - \frac{1}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

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