cos 2xdx
(sinx+cosx)^
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∫(sinx+cosx)2cos2xdx
=∫(sinx+cosx)2cos2x−sin2xdx [∵cos2θ=cos2θ−sin2θ]
=∫(sinx+cosx)2(cosx−sinx)(cosx+sinx)(cosx+sinx)dx
=∫sinx+cosxcosx−sinxdx
Let sinx+cosx=t
(cosx−sinx)dx=dt
∫tdt
lnt+C
ln(sinx+cosx)+C
∴∫(sinx+cosx)2cos2xdx=ln(sinx+cosx)+
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