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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 5 Maths Textbook Solution.

Answers (1)

Answer: \frac{-1}{2} \log \left|2 \cos x+\sqrt{4 \cos ^{2} x-1}\right|+c

Hint Let 2 \cos x=t

Given: \int \frac{\sin x}{\sqrt{4 \cos ^{2} x-1}} d x

Explanation:

            \int \frac{\sin x}{\sqrt{4 \cos ^{2} x-1}} d x        ..........(1)

            Let 2 \cos x=t

            -2 \sin x d x=d t

            \sin x d x=\frac{-1}{2} d t                            (Differentiate w.r.t to t)

Put in (1) we get

                \frac{-1}{2} \int \frac{d t}{\sqrt{t^{2}-1}}                                    

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

                =\frac{-1}{2} \log \left|2 \cos x+\sqrt{4 \cos ^{2} x-1}\right|+c

 

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