Mathematics

# Solve:$\int \displaystyle \dfrac{\sin 2x}{\sin^4x+ \cos^4 x}dx$

##### SOLUTION
Given the integral,
$\int { \dfrac { \sin { 2x } }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } } } dx\\ =\int { \dfrac { 2\sin { x } \cos { x } }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } } } dx\\ =\int { \dfrac { \dfrac { 2\sin { x } \cos { x } }{ \cos ^{ 4 }{ x } } }{ \dfrac { \sin ^{ 4 }{ x } }{ \cos ^{ 4 }{ x } } +\dfrac { \cos ^{ 4 }{ x } }{ \cos ^{ 4 }{ x } } } } dx\\ =\int { \sec ^{ 2 }{ x } \dfrac { 2\tan { x } }{ \tan ^{ 4 }{ x } +1 } } dx$
Let us assume,
$u=\tan { x } \\ \Rightarrow \dfrac { du }{ dx } =\tan { x } \\ \Rightarrow du=\tan { x } dx$
Substituting the values of $u$ and $du$ we get,
$\int { \sec ^{ 2 }{ x } \dfrac { 2\tan { x } }{ \tan ^{ 4 }{ x } +1 } } dx\\ =\int { \dfrac { 2u }{ { u }^{ 4 }+1 } } du$
Again let us assume,
$v={ u }^{ 2 }\\ \Rightarrow \dfrac { dv }{ du } =2u\\ \Rightarrow dv=2udu$
Substituting the values of $v$ and $dv$ we get,
$\int { \dfrac { 2u }{ { u }^{ 4 }+1 } } du\\ =\int { \dfrac { 1 }{ { v }^{ 2 }+1 } } dv\\ =\tan ^{ -1 }{ v } +C\\ =\tan ^{ -1 }{ { u }^{ 2 } } +C\\ =\tan ^{ -1 }{ (\tan ^{ 2 }{ x } ) } +C\\ \therefore \int { \dfrac { \sin { 2x } }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } } } dx=\tan ^{ -1 }{ (\tan ^{ 2 }{ x } ) } +C.$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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