Question

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

Answer 1

Let $\frac{\cos x}{5-3\cos x}=l(5-3\cos x)+m\left(3\sin x\right)+n$
Comparing coefficients of $\cos x,\sin x$ and constant, we get
$-3l=1,3m=0$ and $5l+n=0$
$\Rightarrow l=-1/3,m=0$ and $n=5/3$
Therefore
$I=l\int \frac{5-3\cos x}{5-3\cos x}dx+0+n\int \frac{dx}{5-3cosx}$
$=-\frac{1}{3}\int dx+\frac{5}{3}\int \frac{dx}{5-3[{\cos }^2\left(x/2\right)-si{n}^2\left(x/2\right)]}$
$=-\frac{x}{3}+\frac{5}{3}\int \frac{{\sec }^2\left(x/2\right)dx}{5[1+{\tan }^2\left(x/2\right)]-3}(1-{\tan }^2\left(x/2\right))$
Substitute $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
Hence
$l=-\frac{1}{3}x+\frac{5}{3}\int \frac{2dt}{8t^2+2}=-\frac{1}{3}x+\frac{5}{3}\int \frac{dt}{4t^2+1}$
$=-\frac{1}{3}x+\frac{5}{3.4}\int \frac{dt}{t^2+{\left(\frac{1}{2}\right)}^2}=-\frac{1}{3}x+\frac{5}{12}.\frac{1}{1/2}{\tan }^{-1}\frac{t}{1/2}$
$=-\frac{1}{3}x+\frac{5}{6}{\tan }^{-1}\left[2\tan \frac{x}{2}\right]$