# Tricky Integration Problem $\int _{ \frac { \pi }{ 3 } }^{ \frac { \pi }{ 2 } }{ 64\sin ^{ 2 }{ tcostdt } }$ $u=sint\quad \therefore \quad { u }^{ 2 }=\sin ^{ 2 }{ t } \\ \\ If\quad u=sint,\quad \frac { du }{ dt } =cost,\quad \therefore \quad du=costdt,\quad \therefore \quad \frac { 1 }{ cost } du=dt\\ \\ When\quad t=\frac { \pi }{ 2 } ,\quad u=sin\left( \frac { \pi }{ 2 } \right) =1.\quad When\quad t=\frac { \pi }{ 3 } ,\quad u=sin\left( \frac { \pi }{ 3 } \right) =\frac { \sqrt { 3 } }{ 2 } .\\ \\ So\quad we\quad must\quad integrate:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } \cdot cost\cdot \frac { 1 }{ cost } du,\\ \\ which\quad is:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } du\quad =\quad { \left[ \frac { 64{ u }^{ 3 } }{ 3 } \right] }_{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }\\ \\ =\left( \frac { 64{ \left( 1 \right) }^{ 3 } }{ 3 } \right) -\left( \frac { 64{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 3 } }{ 3 } \right) =\frac { 64 }{ 3 } -8\sqrt { 3 }$