#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 17 Maths Textbook Solution.

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

Given:

$\int \frac{1}{1-\cos x-\sin x} d x$

Hint:

Using partial fraction and $\int \frac{1}{x} d x$.

Explanation:

Let $\mathrm{I}=\int \frac{1}{1-\cos x-\sin x} d x$

\begin{aligned} &\text { Put } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan \frac{x^{x}}{2}} ; \sin x=\frac{2 \tan _{2}^{x}}{1+\tan ^{\frac{x}{2}}}\\ &\therefore I=\int \frac{1}{1-\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)-\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x \end{aligned}

\begin{aligned} &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-1+\tan ^{2}-2 \tan \frac{x}{2}} d x \\ =& \int \frac{\sec ^{2} \frac{x}{2}}{2 \tan ^{2} \frac{x}{2}-2 \tan \frac{x}{2}} d x \\ =& \int \frac{\sec ^{2} \frac{x}{2} \cdot \frac{1}{2}}{\tan \frac{x}{2}\left(\tan _{2}^{x}-1\right)} d x \quad\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right] \end{aligned}

$=\int \frac{d t}{t(t-1)}$                                                        $\text { [ Put } \left.\tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t\right]$

Now,

$\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$

Multiplying by t (t - 1)

$\Rightarrow 1=A(t-1)+B(t)$

Putting t = 1

\begin{aligned} &1=A(0-1)+B(0) \Rightarrow A=-1 \\ &\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1} \\ &\therefore \int \frac{1}{t(t-1)} d t=-\int \frac{1}{t} d t+\int \frac{1}{t-1} d t \end{aligned}

\begin{aligned} &=-\log |t|+\log |t-1|+C \\ &=-\log \left|\tan \frac{x}{2}\right|+\log \left|\tan \frac{x}{2}-1\right|+C \\ &=\log \left|\frac{\tan _{\frac{x}{2}}-1}{\tan _{2}^{\frac{x}{x}}}\right|+C \\ &=\log \left|1-\cot \frac{x}{2}\right|+C \end{aligned}$\left[\because \cot x=\frac{1}{\tan x}\right]$