Let f(x)=2sin2x−1cosx+cosx(2sinx+1)(1+sinx) then ∫ex(f(x)+f′(x))dx where c is the constant of integration
A
extanx+C
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B
excotx+C
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C
excosec2x+C
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D
exsec2x+C
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Solution
The correct option is Cextanx+C We can write f(x)=2sin3x+2sin2x−sinx−1−2sin3x−sin2x+2sinx+1cosx(1+sinx) =sinx(1+sinx)cosx(1+sinx)=tanx We know,∫ex(f(x)+f′(x))=exf(x)+c=extanx+c