How do evaluate the integral below?

Question

#int(cosx+sinx)^2 cos2x dx#

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Answer 1 · 2017-10-04T15:18:52

#(1+sin2x)^2/4+C#

1) I expanded #(cosx+sinx)^2# as #1+sin2x#.

2) I used #u=1+sin2x# and #du=2cos2x*dx# transform.

3) I took integral and made inverse transform.

#int (cosx+sinx)^2*cos2x*dx#

=#int (1+sin2x)*cos2x*dx#

=#int 1/2*(1+sin2x)*2cos2x*dx#

=#int 1/2*u*du#

=#u^2/4+C#

=#(1+sin2x)^2/4+C#