Question

$\int_{}^{}\frac{\mathbf{dx}}{\mathbf{1}\mathbf{-}\mathbf{\cos}\mathbf{x}\mathbf{-}\mathbf{\sin}\mathbf{x}}\mathbf{=}$

• A. $\log\left| 1 + \cot\frac{x}{2} \right| + c$
• B. $\log\left| 1 - \tan\frac{x}{2} \right| + c$
• C. $\log\left| 1 - \cot\frac{x}{2} \right| + c$
• D. $\log\left| 1 + \tan\frac{x}{2} \right| + c$

Answer 1

Answer: (C)

$\log\left| 1 - \cot\frac{x}{2} \right| + c$

Answer 2

Given $I = \int_{}^{}\frac{dx}{1 - \cos x - \sin x}$

$I = \int \frac { d x } { 1 - \frac { \left( 1 - \tan ^ { 2 } x / 2 \right) } { \left( 1 + \tan ^ { 2 } x / 2 \right) } - \frac { 2 \tan x / 2 } { 1 + \tan ^ { 2 } x / 2 } }$ $\Rightarrow I = \int_{}^{}\frac{\sec^{2}x/2.dx}{1 + \tan^{2}x/2 - 1 + \tan^{2}x/2 - 2\tan x/2}I = \int_{}^{}\frac{\sec^{2}x/2.dx}{2\tan^{2}x/2 - 2\tan x/2}$ $\Rightarrow \int_{}^{}\frac{1/2.\sec^{2}x/2.dx}{\tan^{2}x/2 - \tan x/2}$

Put $\tan x/2 = t$ $\Rightarrow \frac{1}{2}\sec^{2}x/2.dx = dt$ therefore $I = \int_{}^{}\frac{dt}{t^{2} - t}$ $\Rightarrow I = \int_{}^{}\frac{dt}{t(t - 1)}$ $\Rightarrow \int_{}^{}\left\lbrack - \frac{1}{t} + \frac{1}{t - 1} \right\rbrack dt \Rightarrow I = \int_{}^{}\frac{dt}{t - 1} - \int_{}^{}\frac{dt}{t}$

$I = \log(t - 1) - \log t + c$

$\Rightarrow I = \log\left| \frac{t - 1}{t} \right| + c$

$\Rightarrow I = \log\left| \frac{\tan x/2 - 1}{\tan x/2} \right| + c$ $\Rightarrow I = \log\left| 1 - \cot x/2 \right| + c$