Let,& I=∫ex[1+cos2x2+sin2x]dx
We know that~sin2x=2sinxcosx and cos2x=2cos2x−1
Using the above formulas we can write
I=∫ex(2cos2x2+2sinxcosx)dx
I=∫ex(2cos2x2(1+sinxcosx))dx
I=∫ex(cos2x1+cos2xsinxcosx)dx
I=∫ex(tanx+sec2x)dx
We know that~∫ex(f(x)+f′(x))dx=ex⋅f+C
Using the above formula we can write
I=extanx+C