Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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