#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 23 Maths Textbook Solution.

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

Given:

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

Hint:

To solve this equation we use sin x formula and substitute method.

Solution:

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

$\left[\sin 2 x=(\cos x+\sin x)^{2}-1, \quad t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x\right.$

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

Let  $t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x$

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

$=-\log \left|t+\sqrt{t^{2}-1}\right|+c$

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

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