${\displaystyle \left[-x\cos x\right]_{0}^{\frac {\pi }{2}}+\int _{0}^{\frac {\pi }{2}}\cos =\left[\sin x-x\cos x\right]_{0}^{\frac {\pi }{2}}=1}$

${\displaystyle \left[t^{2}\sin t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}t\sin t\,\mathrm {d} t=\left[t^{2}\sin t+2t\cos t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}\cos =\left[t^{2}\sin t+2t\cos t-2\sin t\right]_{0}^{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}\left({\frac {\pi ^{2}}{16}}+{\frac {\pi }{2}}-2\right)}$.

${\displaystyle \left[-{\frac {t\cos 3t}{3}}\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\cos 3t}{3}}\,\mathrm {d} t=\left[-{\frac {t\cos 3t}{3}}+{\frac {\sin 3t}{9}}\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3{\sqrt {2}}}}\left({\frac {\pi }{4}}+{\frac {1}{3}}\right)}$.

${\displaystyle \left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\int _{0}^{2\pi }t\cos 2t\,\mathrm {d} t=\left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\left[{\frac {t\sin 2t}{2}}\right]_{0}^{2\pi }-\int _{0}^{2\pi }{\frac {\sin 2t}{2}}\,\mathrm {d} t=-2\pi ^{2}+0-0=-2\pi ^{2}}$.

${\displaystyle \left[x\tan x\right]_{0}^{\frac {\pi }{3}}-\int _{0}^{\frac {\pi }{3}}\tan =\left[x\tan x+\ln \cos x\right]_{0}^{\pi /3}={\frac {\pi {\sqrt {3}}}{3}}-\ln 2}$.

${\displaystyle \left[\ln \tan \right]_{\pi /6}^{\pi /3}=\ln 3}$.

${\displaystyle \int _{0}^{\frac {\pi }{2}}\left(x{\frac {1+\cos 2x}{2}}+{\frac {3}{8}}-{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}\right)\,\mathrm {d} x=}$

${\displaystyle \left[x\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\right]_{0}^{\frac {\pi }{2}}-\int _{0}^{\frac {\pi }{2}}\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\,\mathrm {d} x+\left[{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}=}$

${\displaystyle \left[{\frac {x^{2}}{4}}+{\frac {x\sin 2x}{4}}+{\frac {\cos 2x}{8}}+{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi ^{2}}{16}}-{\frac {1}{4}}+{\frac {3\pi }{16}}}$.

${\displaystyle \int _{1}^{0}-2x^{2}\,\mathrm {d} x+2\int _{0}^{\frac {\pi }{2}}\left(1-\cos 2t\right)\,\mathrm {d} t={\frac {2}{3}}+\pi }$.