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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.

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

Answers (2)

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

View full answer

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