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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.

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

Answers (1)

Answer:

\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                                                                                      

 

Posted by

infoexpert21

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