#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 17 Maths Textbook Solution.

Answer: $-\log |(\sin x+\cos x)+\sqrt{\sin 2 x}|+c$

Hint  $\sin ^{2} x+\cos ^{2} x=1$

Given:$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$

Explanation:

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

$\int \frac{\sin x-\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x-1}} d x$                    $\left[\begin{array}{l} \sin ^{2} x+\cos ^{2} x=1 \\ \sin 2 x=2 \sin x \cdot \cos x \end{array}\right]$

$\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^{2}-1}} d x$                     ...........(1)            $\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

Let $\sin x+\cos x=t$

$(\cos x-\sin x) d x=d t$

$-(\sin x-\cos x) d x=d t$

From (1) we have

$-\int \frac{d t}{\sqrt{t^{2}-1}}$

$=-\log \left|t+\sqrt{t^{2}-1}\right|+c$                                            $\left[\because \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right]$

$=-\log \left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^{2}-1}\right|+c$

$=-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+c$