Question

Evaluate $\int \frac{dx}{\sin x+\tan x}$ =$\frac{1}{m}\ln |\frac{(cosx-1)}{(cosx+1)}|-\frac{1}{2}\cdot \frac{1}{(cosx+1)}+C$ then $m=$

Answer 1

$\int \frac{1}{\sin x+\tan x}dx\Rightarrow$ put $\tan x=\frac{\sin x}{\cos x}$

${\sin }^{r}x=1-{\cos }^{r}x$

$=\int -\frac{\cos x}{(\cos x-1)(\cos x+1{)}^r}\sin xdx$

Let $u=\cos x\Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=-\frac{du}{\sin x}$

$=\int \frac{u}{(u-1)(u+1{)}^r}du$

Perform partial fractions

$=\int (-\frac{1}{4(u+1)}+\frac{1}{2(u+1{)}^r}+\frac{1}{4(u-1)})du$

$=\frac{-1}{4}\int \frac{du}{u+1}+\frac{1}{2}\int \frac{du}{(u+1{)}^r}+\frac{1}{4}\int \frac{du}{u-1}$

$=\frac{-1}{4}\ln (u+1)+\frac{1}{2}\left[\frac{-1}{u+1}\right)+\left(\frac{1}{4}\right)\ln (u-1)$

$=-\frac{\ln (\cos x+1)}{4}+\frac{\ln (\cos x-1)}{4}-\frac{1}{2(\cos x+1)}+c$

Compare with $\frac{1}{m}\ln \left|\frac{\cos x-1}{\cos x+1}\right|-\frac{1}{2(\cos x+1)}+c$

we get $m=4$