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 $$k$$.


ANSWER

5


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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