Express cos2x in the form a*cos^2(x) + b and hence show that the integral of cos^2(x) between 0 and pi/2 is equal to pi/a.

Apply the double angle formula to cos2x to yield the requested result.
cos2x = 2cos^2(x) - 1
Spot that the question asks us to prove the value of cos^2(x) when integrated, and that we can move the variables in the above equation to have cos^2(x) on its own.
cos^2(x) = (1/2)*(cos2x +1)
Now we can integrate the the equation between 0 and pi, and we should get the right hand side equal to pi/4.
[ (1/4)*sin2x + x/2 ] from 0 to pi/2
substituting pi/2 into the above equation gives pi/4. Substituting 0 into the above equation gives 0.
So we get pi/4 - 0 = pi/4

LP

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