We have,
I=∫cos2x(cosx+sinx)2dx
I=∫cos2xcos2x+sin2x+2sinxcosxdx
I=∫cos2x1+sin2xdx
Let t=1+sin2x
dtdx=0+2cos2x
dt2=cos2xdx
Therefore,
I=12∫dtt
I=12ln(t)+C
Put the value of t, we get
I=12ln(1+sin2x)+C
Hence, this is the answer.
Find the integrals of the functions. ∫cos2x(cosx+sinx)2dx.