Question

Solve: $\int \frac{\cot x}{1+\cot x+{\cot }^{2}x}dx$

• A. $x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+c$
• B. $x+\frac{2}{\sqrt{3}}{\cot }^{-1}\left(\frac{2\cot x+1}{\sqrt{3}}\right)+c$
• C. $x+\frac{\sqrt{3}}{2}{\cot }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+c$
• D. $x^2+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\cot x+1}{\sqrt{3}}\right)+c$

Answer 1

The correct option is A $x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+c$

$\int \frac{\cot x}{1+\cot x+\cot x^{2}x}dx$

$=\int \frac{\cot x/\sin x}{1+\frac{\cos x}{\sin x}}+\frac{{\cos }^{2}n}{{\sin }^{2}n}dxdn$

$=\int \frac{\cos x/\sin x}{\frac{{\sin }^{2}x+\cos x\sin x+{\cos }^{2}x}{{\sin }^{2}x}}dn$

$\int \frac{\cos x\sin x}{{\sin }^2+\cos n\sin x+{\cos }^{2}n}dx$

$\int \frac{\sin \cos x}{1+\sin x\cos x}dn$

$=\int \frac{1+\sin x\cos x-1}{1-\sin x\cos x}dn$

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

$\int dx-\int \frac{2}{2+\sin x}dn$

$x-\int \frac{2}{2+\frac{2\tan x}{1+{\tan }^{2}x}}dx$

$=x-\int \frac{{\sec }^{2}x}{(1+{\tan }^{2}x+\tan x)}dx$

$x-\int \frac{dt}{1+t+t^2}$ let $\tan x=t$ ${\sec }^{2}xdn=dt$

$x-\int \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$=x-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t+1/2}{\sqrt{3}/2}\right)+C$

$I=x\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C$