#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 11 Maths Textbook Solution.

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

Given:

$\int \frac{1}{1+\tan x} d x$

Hint:

You must know about the derivation of sin x and cos x and use $\int \frac{d x}{x}$.

Explanation:

Let $I=\int \frac{1}{1+\tan x} d x$

$=\frac{1}{1+\tan x}$

$=\frac{1}{1+\frac{\sin x}{\cos x}}$                                                                $\left[\because \tan x=\frac{\sin x}{\cos x}\right]$

$=\frac{\cos x}{\cos x+\sin x}$

$=\frac{2 \cos x}{2(\cos x+\sin x)}$

$=\frac{(\cos x+\sin x)+(\cos x-\sin x)}{2(\cos x+\sin x)}$                                  [Special step]

\begin{aligned} &\quad=\frac{1}{2}\left[1+\frac{\cos x-\sin x}{\cos x+\sin x}\right] \\ &\therefore I_{2}=\int \frac{1}{1+\tan x} d x \\ &\quad=\int \frac{1}{2} d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \\ &\therefore I_{2}=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{d t}{t} \end{aligned}                                  $\left[\operatorname{Put} \cos x+\sin x=t \text { in } I_{2}\right]$

$[(-\sin x+\cos x) d x=d t \Rightarrow(\cos x-\sin x) d x=d t]$

$=\frac{1}{2} x+\frac{1}{2} \log |t|+C$

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