Question

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

विकल्प

• \[\log\left| 1 + \cot\frac{x}{2} \right| + C\]
• \[\log\left| 1 - \tan\frac{x}{2} \right| + C\]
• \[\log\left| 1 - \cot\frac{x}{2} \right| + C\]
• \[\log\left| 1 + \tan\frac{x}{2} \right| + C\]

Answer 1

\[\log\left| 1 - \cot\frac{x}{2} \right| + C\]

 

 

\[\text{Let }I = \int\frac{dx}{1 - \cos x - \sin x}\]

\[ = \int\frac{dx}{1 - \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) - \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}\]

\[ = \int\frac{\left( 1 + \tan^2 \frac{x}{2} \right) dx}{\left( 1 + \tan^2 \frac{x}{2} \right) - \left( 1 - \tan^2 \frac{x}{2} \right) - 2 \tan \frac{x}{2}}\]

\[ = \int\frac{\sec^2 \frac{x}{2} dx}{2 \tan^2 \frac{x}{2} - 2 \tan \frac{x}{2}}\]

\[ = \frac{1}{2}\int\frac{\sec^2 \frac{x}{2} dx}{\tan^2 \frac{x}{2} - \tan \frac{x}{2}}\]
\[\text{Putting }\tan \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) dx = dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right) dx = 2dt\]
\[ \therefore I = \frac{1}{2}\int\frac{2dt}{t^2 - t}\]
\[ = \int\frac{dt}{t^2 - t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = \int\frac{dt}{\left( t - \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{2 \times \frac{1}{2}} \ln \left| \frac{t - \frac{1}{2} - \frac{1}{2}}{t - \frac{1}{2} + \frac{1}{2}} \right| + C ............\left( \because \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left| \frac{x - a}{x + a} \right| + C \right)\]
\[ = \ln \left| \frac{t - 1}{t} \right| + C\]
\[ = \ln \left| 1 - \frac{1}{t} \right| + C\]
\[ = \ln \left| 1 - \frac{1}{\tan \frac{x}{2}} \right| + C ..........\left( \because t = \tan \frac{x}{2} \right)\]
\[ = \ln \left| 1 - \cot \frac{x}{2} \right| + C\]