#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 16 Maths Textbook Solution.

Answer:$\log \left(\sin x+\frac{1}{2}\right)+\sqrt{\sin ^{2} x+\sin x} \mid+c$

Hint  Let $\sin x=t$

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

Explanation:

$\int \sqrt{\cos e c x-1} d x$

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

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

Multiply and Divide with (1 + sin x)

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

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

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

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

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

Let $\sin x=t$

$\operatorname{Cos} x d x=d t$

From (1)

$\int \frac{d t}{\sqrt{t(1+t)}}$

$=\int \frac{d t}{\sqrt{t^{2}+t+\frac{1}{4}-\frac{1}{4}}}$

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

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

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