How do you find the integral of #int (1 + cos x)^2 dx#?

Question
Answer 1

#int(1+cosx)^2dx=1/4(6x+8sinx+sin2x)+"c"#

#int(1+cos^2x) dx=int1+2cosx+cos^2xdx#
Then use the reduction formula on #cos^2x# to get it into an integrable form
#int1+2cosx+cos^2xdx=int1+2cosx+1/2+1/2cos2xdx=int3/2+2cosx+1/2cos2xdx#
#int3/2+2cosx+1/2cos2x=3/2x+2sinx+1/4sin2x+"c"=1/4(6x+8sinx+sin2x)+"c"#
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Answer 2
#int(1+cos x)^2 dx# =#int(1+2cos x +cos^2 x) dx# =#int(1+2cos x +(1/2cos 2x - 1/2) dx# * =#int(1/2 + 2cos x + 1/2cos 2x) dx# =#1/2x +2sin x +1/4sin 2x +c#
*#1 + 2cos^2 x = cos 2x# #:. cos^2 x = 1/2cos 2x -1/2#
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