Question

$\int\frac{dx}{1-cos x -sin x} $ is equal to

• 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)

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

put $cos x = \frac{1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}} and sin x= \frac{2tan \frac{x}{2}}{1+tan ^{2 } \frac{x}{2}}$

$ \therefore I= \int\frac{dx}{1-\left(\frac{1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}}\right)- \frac{2tan \frac{x}{2}}{1+tan ^{ 2} \frac{x}{2}}} $

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

Put $tan \frac{x}{2}=t \Rightarrow \frac{1}{2} sec^{2} \frac{x}{2 }dx = dt $

$\therefore I = \int \frac{dt}{t^{2} -t} = \left[\frac{1}{t-1}-\frac{1}{t}\right]\int+C $

$=log\left(t-1\right) -logt + C = log\left|\frac{t-1}{t} +C\right| $

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