Verified answer

the trick here is to use some trig substitutions

start with:

cos(2x)=cos^2(x)-sin^2(x)

along with:

cos^2(x)+sin^2(x)=1

then we have that

cos(2x)=cos^2(x)-[1-cos^2(x)]

=2cos^2(x)-1

which leads to:

cos^2(x)=1/2[1-cos(2x)]

integrating gives you:

Integral[cos^2(x) dx]=1/2 Integral[1-cos(2x) dx]

evaluating should give you the value pi

Integral of cos^2(Φ) dΦ over interval 0 to 2π = (1/2)2π = π

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Ideas: cos^2(Φ) = (1/2)(1+cos2Φ), and the period of cos2Φ is π.

use the fact that cos(2Φ) = 2cos^2(Φ) - 1

So just replace:

cos^2(Φ) = (cos(2Φ) + 1) / 2

Integral of this is (1/2) [ (1/2)sin(2Φ) + Φ ]

Just substitute 0 and 2π...