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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 9 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{2} \log \left|\sin 2 x+\sqrt{\sin ^{2} 2 x+8}\right|+c

Hint      \sin 2 x=t

Given: \int \frac{\cos 2 x}{\sqrt{\sin ^{2} 2 x+8}} d x

Explanation:

            \int \frac{\cos 2 x}{\sqrt{\sin ^{2} 2 x+8}} d x                    ...........(1)

          Let

           \sin 2 x=t

           2 \cos 2 x d x=d t

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

 Put in (1), we have

                \frac{1}{2} \int \frac{d t}{\sqrt{t^{2}+8}}

              =\frac{1}{2} \log \left|t+\sqrt{t^{2}+8}\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]

               =\frac{1}{2} \log \left|\sin 2 x+\sqrt{\sin ^{2} 2 x+8}\right|+c

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