Question

$\int \frac{1}{\tan x+\cot x+\sec x+co\sec x}dx$, taking $c=0$, we get $I=\frac{1}{k}(\sin x-\cos x-x)$
Find $k$

Answer 1

Change in terms of $\sin x$ and $\cos x$ $I=\int \frac{\sin x\cos x}{1+\sin x+\cos x}dx=\int \frac{\sin xdx}{\sec x+\tan x+1}.$
Multiply above and below by $(1+\tan x)-\sec x$ $I=\int \frac{\sin x(1+\tan x-\sec x)}{(1+\tan x)-{\sec }^{2}x}dx$
$D^r=(1+{\tan }^{2}x)-{\sec }^{2}x+2\tan x=0+2\sin x/\cos x$
$\therefore I=\frac{1}{2}\int \cos x(1+\frac{\sin x}{\cos x}-\frac{1}{\cos x})dx\mid =\frac{1}{2}\int (\cos x+\sin x-1)dx$ $=\frac{1}{2}(\sin x-\cos x-x)$
Ans: 2