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Find the average value of the function \sin ^{6}(t)\cos ^{3}(t) on the interval \left[0,{\frac  {\pi }{2}}\right].

For the method to do this integral see the section on Substitution and the section on Trigonometric Integrals. The average value of this function on the interval \left[0,{\frac  {\pi }{2}}\right] is

{\frac  {1}{{\frac  {\pi }{2}}-0}}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{6}(t)\cos ^{3}(t)\,dt={\frac  {2}{\pi }}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{6}t(1-\sin ^{2}t)\cos t\,dt={\frac  {2}{\pi }}\int _{{0}}^{{{\frac  {\pi }{2}}}}(\sin ^{6}t-\sin ^{8}t)\cos t\,dt\,, which can be integrated at sight,

={\frac  {2}{\pi }}\left[{\frac  {\sin ^{7}t}{7}}-{\frac  {\sin ^{9}t}{9}}\right]_{{0}}^{{{\frac  {\pi }{2}}}}={\frac  {2}{\pi }}\left[{\frac  {1}{7}}-{\frac  {1}{9}}\right]={\frac  {4}{63\pi }}\,

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