Question

If $\int \frac{1}{\tan x+\mathrm{co}tx}$ dx = k cos 2x + C, then k = ___________________.

Answer 1

$\int \frac{1}{\tan x+\cot x}dx$

$=\int \frac{1}{{\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{\cos x}{\sin x}}}dx$

$=\int \frac{1}{{\displaystyle \frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}}}dx$

$=\int \sin x\cos xdx$

$=\frac{1}{2}\int \sin 2xdx\left(\sin 2A=2\sin A\cos A\right)$

$=\frac{1}{2}\times \left(-\frac{\cos 2x}{2}\right)+C$

$=-\frac{1}{4}\cos 2x+C$

Comparing with the given expression, we get

$k=-\frac{1}{4}$


If $\int \frac{1}{\tan x+\cot x}dx$ = k cos 2x + C, then k = $\underline{-\frac{1}{4}}$.