I do not know what this question is about but
\(cos2\theta=cos^2\theta-sin^2\theta \\ cos2\theta=cos^2\theta-(1-cos^2\theta) \\ \frac{cos2\theta+1}{2}=cos^2\theta\\\)
so
\(\displaystyle\int_0^{\pi/4} cos^2\theta\;d\theta \\ =\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)+1}{2}\;d\theta \\ =\frac{1}{2}\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)}{1}+1\;d\theta \\ =\frac{1}{2}\left[ \frac{sin(2\theta)}{2}+\theta\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ sin(2\theta)+2\theta\;\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ (sin(\pi)+\pi)-(0)\;\right ]\\ =\frac{\pi}{4}\)
I don't know if that helps or not.
+118727 
