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Question

Evaluate:
(xcosx+sinxxsinx)dx

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Solution

Consider the given integral.
I=(xcosx+sinxxsinx)dx

Let
t=xsinx
dtdx=xcosx+sinx×1

dt=(xcosx+sinx)dx

Therefore,
I=1tdt
I=ln t+C

On putting the value of the t, we get
I=ln (xsinx)+C

Hence, this is the answer.

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