Mathematics

$$\int_{}^{} {{{\tan }^3}2x\sec 2x{\rm{ dx}}} $$ is equal to


ANSWER

$$\dfrac{1}{6}{\sec ^3}2x - \frac{1}{2}\sec 2x + c$$


SOLUTION
$$\int { \tan ^{ 3 }{ 2x } \sec { 2x } dx } $$
$$\int { \dfrac { \sin ^{ 3 }{ 2x }  }{ \cos ^{ 4 }{ 2x }  }  } dx$$
Let $$\cos 2x=t$$
On differentiating, we have
$$=-2\sin 2x dx=dt$$
$$=\int { \dfrac { -\sin ^{ 2 }{ 2x }  }{ 2{ t }^{ 4 } }  } dt$$
$$=-\int \dfrac{1}{2}\dfrac { \left( 1-\cos ^{ 2 }{ 2x }  \right)  }{ { t }^{ 4 } } dt$$
$$=-\dfrac{1}{2}\left[ \int { \dfrac { 1 }{ { t }^{ 2 } } -\int { \dfrac { 1 }{ { t }^{ 4 } }  }  } dt \right] $$
$$=\dfrac { { t }^{ -2+1 } }{ -2+1 } -\dfrac { { t }^{ -4+1 } }{ -4+1 } +c$$
$$=\dfrac { t }{ 2 } \left[ -\dfrac { 1 }{ t } +\dfrac { 1 }{ 3{ t }^{ 3 } } +c \right] $$
$$=\dfrac { 1 }{ 2 } \left[ -\dfrac { 1 }{ 3{ t }^{ 3 } } +\dfrac { 1 }{ t } +c \right] $$
$$=\dfrac { 1 }{ 2 } \left[ -\dfrac { 1 }{ 3\cos ^{ 2 }{ 2x }  } +\dfrac { 1 }{ \cos { 2x }  } +c \right] $$
$$=\dfrac { 1 }{ 6 } \sec ^{ 3 }{ 2x } -\dfrac { 1 }{ 2 } \sec { 2x } +\dfrac { c }{ 2 } $$
Hence, the answer is $$\dfrac { 1 }{ 6 } \sec ^{ 3 }{ 2x } -\dfrac { 1 }{ 2 } \sec { 2x } +\dfrac { c }{ 2 }.$$
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Single Correct Medium Published on 17th 09, 2020
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