∫exsinx+2cosxsinxdx=
excosx+c
exsinx+c
exsin2x+c
ex(cosx+sinx)+c
Explanation for the correct Option:
Integrating using substitution:
Given
∫exsinx+2cosxsinxdx=∫exsin2xdx+2exsinxcosxdx=∫exsin2xdx+2sinxcosxdx
Let's take fx=sinx2f'x=2sinxcosx
Substituting the values, we get
=∫exf(x)+f'(x)dx=exf(x)+c∵∫exf(x)+f'(x)dx=exf(x)+c=exsin2x+c∵f(x)=sin2x
Hence, the correct answer is option (C).