#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 49 Maths Textbook Solution.

$I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c$

Given:

$\int \sqrt{\operatorname{cosec} x-1} d x$

Hint:

To solve this statement we have to convert cosec into sin and then we apply standard formula.

Solution:

$\int \sqrt{\operatorname{cosec} x-1} d x$

$I=\int \sqrt{\frac{1}{\sin x}-1} d x$

$I=\int \sqrt{\frac{1-\sin x}{\sin x}}-d x$

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

$I=\int \frac{\sqrt{1-\sin ^{2} x}}{\sqrt{\sin x(1+\sin x)}} d x$

$I=\int \frac{\sqrt{\cos ^{2} x}}{\sqrt{\sin x(1+\sin x)}} d x$

$I=\int \frac{\cos x}{\sqrt{\sin x(1+\sin x)}} d x$

sin x = t ,

dt = cos x dx

$\therefore I=\int \frac{d t}{\sqrt{t}(t+1)}=\int \frac{d t}{\sqrt{t^{2}+t}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}$

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

$I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c$