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

${\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x=\left[\tan x-x\right]_{-\pi /4}^{\pi /4}=2-{\frac {\pi }{2}}}$.

${\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\tan x}{\cos x}}\,\mathrm {d} x=0}$, par imparité.

${\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t&=\int _{0}^{\frac {\pi }{3}}\left(2\cos ^{2}t-1\right)\left(\cos ^{2}t-1\right)(-\sin t)\,\mathrm {d} t\\&=\int _{1}^{1/2}\left(2x^{2}-1\right)\left(x^{2}-1\right)\,\mathrm {d} x\\&=\int _{1}^{1/2}\left(2x^{4}-3x^{2}+1\right)\,\mathrm {d} x\\&=\left[{\frac {2x^{5}}{5}}-x^{3}+x\right]_{1}^{1/2}\\&={\frac {-1}{80}}.\end{aligned}}}$

${\displaystyle 2\cos a\cos b=\cos(a+b)+\cos(a-b)}$ donc ${\displaystyle \cos t\cos 2t\cos 3t={\frac {1}{2}}(\cos 3t+\cos t)\cos 3t={\frac {1}{4}}(1+\cos 6t+\cos 4t+\cos 2t)}$ et

${\displaystyle {\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}(1+\cos 6t+\cos 4t+\cos 2t)\,\mathrm {d} t={\frac {1}{4}}\left[t+{\frac {\sin 6t}{6}}+{\frac {\sin 4t}{4}}+{\frac {\sin 2t}{2}}\right]_{0}^{\frac {\pi }{2}}=\left[{\frac {t}{4}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{8}}}$.

${\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t=-\int _{1/{\sqrt {2}}}^{-1/{\sqrt {2}}}\left(x^{2}+(1-x^{2})^{2}\right)\,\mathrm {d} x=\left[{\frac {x^{5}}{5}}-{\frac {x^{3}}{3}}+x\right]_{-1/{\sqrt {2}}}^{1/{\sqrt {2}}}={\sqrt {2}}\left({\frac {1}{20}}-{\frac {1}{6}}+1\right)={\frac {53}{60}}}$.

${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t=\int _{0}^{1}{\frac {\mathrm {d} x}{5-3x}}={\frac {-1}{3}}\left[\ln(5-3x)\right]_{0}^{1}={\frac {\ln(5/2)}{3}}}$.

${\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}{\frac {2\sin t}{1+\cos ^{2}{\frac {t}{2}}}}\,\mathrm {d} t=-4\int _{7/4}^{3/4}{\frac {\mathrm {d} x}{x}}=4\ln(7/3)}$.