Mathematics

# $\displaystyle \int \frac{\cos x}{5-3\cos x}dx=-\frac{1}{3}x+\frac{k}{6}\tan ^{-1}\left [ 2\tan \frac{x}{2} \right ].$ Find the value of $k$.

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##### SOLUTION
Let $\displaystyle \frac { \cos x }{ 5-3\cos x } =l\left( 5-3\cos x \right) +m\left( 3\sin x \right) +n$
Comparing coefficients of $\displaystyle \cos x,\sin x$ and constant, we get
$-3l=1,3m=0$ and $5l+n=0$
$\Rightarrow l=-1/3,m=0$ and $n=5/3$
Therefore
$\displaystyle I=l\int \frac { 5-3\cos x }{ 5-3\cos x } dx+0+n\int \frac { dx }{ 5-3cosx }$

$\displaystyle =-\frac { 1 }{ 3 } \int dx+\frac { 5 }{ 3 } \int \frac { dx }{ 5-3\left[ \cos ^{ 2 } \left( x/2 \right) -sin^{ 2 }\left( x/2 \right) \right] }$

$\displaystyle =-\frac { x }{ 3 } +\frac { 5 }{ 3 } \int \frac { \sec ^{ 2 } \left( x/2 \right) dx }{ 5\left[ 1+\tan ^{ 2 } \left( x/2 \right) \right] -3 } \left( 1-\tan ^{ 2 } \left( x/2 \right) \right)$

Substitute $\displaystyle \tan \frac { x }{ 2 } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 } \frac { x }{ 2 } dx=dt$
Hence
$\displaystyle l=-\frac { 1 }{ 3 } x+\frac { 5 }{ 3 } \int \frac { 2dt }{ 8t^{ 2 }+2 } =-\frac { 1 }{ 3 } x+\frac { 5 }{ 3 } \int \frac { dt }{ 4t^{ 2 }+1 }$

$\displaystyle =-\frac { 1 }{ 3 } x+\frac { 5 }{ 3.4 } \int \frac { dt }{ t^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } =-\frac { 1 }{ 3 } x+\frac { 5 }{ 12 } .\frac { 1 }{ 1/2 } \tan ^{ -1 } \frac { t }{ 1/2 }$

$\displaystyle =-\frac { 1 }{ 3 } x+\frac { 5 }{ 6 } \tan ^{ -1 } \left[ 2\tan \frac { x }{ 2 } \right]$

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One Word Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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