#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 28 Maths Textbook Solution.

$\pm \log (\sin x-\cos x)+C$

Given:

$\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}}$

Hint:

You must know about the derivation of sin and cos function and $\int \frac{1}{x} d x$ .

Explanation:

Let $I=\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}} d x$

$=\int \frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x-\sin 2 x}} d x$                                        $\left[\because \sin ^{2} x+\cos ^{2} x=1\right]$

$=\int \frac{\sin x+\cos x}{\sqrt{(\cos x-\sin x)^{2}}} d x$                                                        $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

\begin{aligned} &=\int \frac{\sin x+\cos x}{(\cos x-\sin x)} d x \\ &=\int \frac{d t}{ t} \end{aligned}                                                                      $[\text { Put } \cos x-\sin x=t \Rightarrow(\cos x+\sin x) d x=d t]$

\begin{aligned} &= \log |t|+C \\ &= \log (\cos x-\sin x)+C \end{aligned}

Hence,$\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}}= \log (\cos x-\sin x)+C .$