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Question

I=ex(2+sin2x)(1+cos2x)dx.

A
exsinx.
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B
excosx.
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C
extanx.
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D
excos2x.
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Solution

The correct option is B extanx.
I=ex(2+sin2x)(1+cos2x)dx.
=[ex.22cos2x+ex2sinxcosx2cos2x]dx
exsec2xdx+extanxdx.
Integrate first by parts
I=extanxextanxdx+extanxdx.
=extanx. The last two integrals cancel.
Above is of the form
ex[f(x)+f(x)]dx=exf(x),
where f(x)=tanx and f(x)=sec2x.

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