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

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


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


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


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                                                                                      


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