Question

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

(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

(c) $\log \left|1-\cot \frac{x}{2}\right|+C$
Disclaimer : Here in answer $\log \left|1-\cot \frac{x}{2}\right|+C$ refers to ${\log }_e\left|1-\cot \frac{x}{2}\right|+C\mathrm{or}\ln \left|1-\cot \frac{x}{2}\right|+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{1-\cos x-\sin x} \\ =\int \frac{dx}{1-\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)-{\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}} \\ =\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)dx}{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)-\left(1-{\tan }^2{\displaystyle \frac{x}{2}}\right)-2\tan {\displaystyle \frac{x}{2}}} \\ =\int \frac{{\sec }^2\mathit{}{\displaystyle \frac{x}{\mathit{2}}}\mathit{}dx}{2{\tan }^2{\displaystyle \frac{x}{2}}-2\tan {\displaystyle \frac{x}{2}}} \\ =\frac{1}{2}\int \frac{{\sec }^2{\displaystyle \frac{x}{2}}dx}{{\tan }^2{\displaystyle \frac{x}{2}}-\tan {\displaystyle \frac{x}{2}}} \\ \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore I=\frac{1}{2}\int \frac{2dt}{t^2-t} \\ =\int \frac{dt}{t^2-t+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{1}{2\times {\displaystyle \frac{1}{2}}}\ln \left|\frac{t-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}{t-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \int \frac{dx}{x^{\mathit{2}}\mathit{-}{a}^{\mathit{2}}}=\frac{1}{2a}\ln \left|\frac{x\mathit{-}a}{x\mathit{+}a}\right|+C\right) \\ =\ln \left|\frac{t-1}{t}\right|+C \\ =\ln \left|1-\frac{1}{t}\right|+C \\ =\ln \left|1-\frac{1}{\tan {\displaystyle \frac{x}{2}}}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t\mathit{=}\tan \mathit{}\frac{x}{\mathit{2}}\right) \\ =\ln \left|1-\cot \frac{x}{2}\right|+C\mathit{} \end{gathered}$$