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Question

Evaluate: ex(2+sin2x)cos2xdx

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Solution

I=ex(2cos2x+2sinxcosxcos2x)dx
=ex(2sec2x+2tanx)dx
=2ex(sec2x+tanx)dx
=2[exsec2xdx+extanxdx]
=2[exsec2xdx{ddxexsec2xdx}dx+extanxdx]
=2[extanxextanxdx+extanxdx]
=2extanx+c

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