If ∫esin2xsinx(cosx+cos3x)dx=f(x)+C, where C is a constant of integration, then f(x) equals
A
12esin2x(3−sin2x)
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B
12esin2x(1−12cos2x)
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C
esin2x(3cos2x+2sin2x)
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D
esin2x(2cos2x+3sin2x)
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Solution
The correct option is A12esin2x(3−sin2x) I=∫esin2xsinx(cosx+cos3x)dx I=∫esin2xsinxcosxdx+∫esin2xsinxcos3xdx Put sin2x=t⇒2sinxcosxdx=dt I=12∫etdt+12∫et(1−t)dt =∫etdt−12∫tetdt =et−12[tet−et]+C =32et−12tet+C =12(3−sin2x)esin2x+C