Question

\[\int\frac{1}{1 + \tan x} dx =\]

विकल्प

• log_e (x + sin x) + C

• log_e (sin x + cos x) + C

• \[2 \sec^2 \frac{x}{2} + C\]

• \[\frac{1}{2}\] [x + log (sin x + cos x)] + C

Answer 1

\[\frac{1}{2}\]  [x + ln (sin x + cos x)] + C

 

\[\text{Let }I = \int\frac{1}{1 + \tan x}dx\]
\[ = \int\frac{1}{1 + \frac{\sin x}{\cos x}}dx\]
\[ = \int\frac{\cos x}{\cos x + \sin x}dx\]
\[ = \frac{1}{2}\int\frac{2 \cos x}{\cos x + \sin x}dx\]
\[ = \frac{1}{2}\int\left[ \frac{\left( \cos x + \sin x \right) + \left( \cos x - \sin x \right)}{\left( \cos x + \sin x \right)} \right]dx\]
\[ = \frac{1}{2}\int\left( \frac{\cos x + \sin x}{\cos x + \sin x} \right)dx + \frac{1}{2}\int\left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)dx\]
\[ = \frac{1}{2}\int dx + \frac{1}{2}\int\left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)dx\]
\[\text{Putting }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]
\[ \therefore I = \frac{1}{2}\int dx + \frac{1}{2}\int\frac{dt}{t}\]
\[ = \frac{x}{2} + \frac{1}{2}\ln \left| t \right| + C\]
\[ = \frac{x}{2} + \frac{1}{2} \ln \left| \cos x + \sin x \right| + C .............\left( \because t = \sin x + \cos x \right)\]
\[ = \frac{1}{2}\left[ x + \ln \left( \sin x + \cos x \right) \right] + C\]