I have to evaluate
$$I = \int_0^{\frac{\pi}{2}}{|\frac{1}{2}-\sin^2x|\mathrm dx}$$
What I've got:
$$\int_0^{\frac{\pi}{2}}{\left|\frac{1}{2}-\sin^2x\right|\mathrm dx}=\int_0^{\frac{\pi}{2}}{\left|\frac{1}{2}-\frac{1-\cos2x}{2}\right|\mathrm dx}=\frac{1}{4}\int_0^{\frac{\pi}{2}}{\left|\cos2x\right|\mathrm d(2x)}$$
The $\cos x$ is positive from $0$ to $\frac{\pi}{2}$ so I guess taking the absolute value is the same as just taking the value.
$$I=\frac{1}{4}\int_0^{\frac{\pi}{2}}{\cos2x\,\mathrm d(2x)}=\frac{1}{4}(\sin{\pi}-\sin{0})+C=C-\frac{1}{4}$$
Is this correct? Thanks in advance!
$$\int_0^{\pi/2}|\cos2x|dx=\int_0^{\pi/4}\cos2x\ dx+\int_{{\pi/4}}^{\pi/2}(-\cos2x)\ dx$$
As $\cos2x\le0$ for $\dfrac\pi2\le2x\le\pi$