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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 48 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 48 Maths Textbook Solution.

Answers (1)

Answer:

I=-\ln \left|(\cos x-1)+\sqrt{\cos ^{2} x-2 \cos x+3}\right|+c

Given:

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

Hint:

To solve this statement we have to use standard formula.

Solution: 

I=\int \frac{\sin x}{\sqrt{\cos ^{2} x-2 \cos x-3}} d x                                [\because \cos x=t, d t=-\sin x d x=>\sin x d x=-d t]

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

I=-\int \frac{d t}{\sqrt{(t-1)^{2}-1-3}}

 

I=-\int \frac{d t}{\sqrt{(t-1)^{2}-4}}

I=-\int \frac{d t}{\sqrt{(t-1)^{2}-(2)^{2}}}                                    \left[\because \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right.

I=-\ln \mid(t-1)+\sqrt{(t-1)^{2}-(2)^{2} \mid+} c

I=-\ln \left|(t-1)+\sqrt{t^{2}-2 t-3}\right|+c

I=-\ln \left|(\cos x-1)+\sqrt{\cos ^{2} x-2 \cos x+3}\right|+c

 

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