#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.18 Question 15 Maths Textbook Solution.

Answer: $\log \left|(\sin x-1)+\sqrt{\sin ^{2} x-2 \sin x-3}\right|+c$

Hint Let $\sin x=t$

Given: $\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x$

Explanation:

$\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x$            ...........(1)

Let $\sin x=t$

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

Put in (1) we have

$\int \frac{d t}{\sqrt{t^{2}-2 t-3}}$

$=\int \frac{d t}{\sqrt{t^{2}-2 t+1-1-3}}$

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

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

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

$=\log \left|(\sin x-1)+\sqrt{\sin ^{2} x-2 \sin x-3}\right|+c$

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Answer: $\log \left|(\sin x-1)+\sqrt{\sin ^{2} x-2 \sin x-3}\right|+c$

Hint Let $\sin x=t$

Given: $\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x$

Explanation:

$\int \frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}} d x$            ...........(1)

Let $\sin x=t$

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

Put in (1) we have

$\int \frac{d t}{\sqrt{t^{2}-2 t-3}}$

$=\int \frac{d t}{\sqrt{t^{2}-2 t+1-1-3}}$

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

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

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

$=\log \left|(\sin x-1)+\sqrt{\sin ^{2} x-2 \sin x-3}\right|+c$