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

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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

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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

Posted by

infoexpert21

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