\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\)
\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)
\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)
\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)
