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Question

sinx(3+2cosx)dx

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Solution

sinx(3+2cosx)dx
(3sinx+2sinxcosx)dx
3sinx+sin2xdx [2sinxcosx=sin2x]
=3(cosx)+(cos2x)2+c
=3cosxcos2x2+c
Hence, the answer is 3cosxcos2x2+c.

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