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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 7 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{3} \sin ^{-1}\left(\frac{3 \log x}{2}\right)+c

Hint:  Let 3 \log x=2

Given: \int \frac{1}{x \sqrt{4-9(\log x)^{2}}} d x

Explanation:

            \int \frac{1}{x \sqrt{4-9(\log x)^{2}}} d x   ........(1)

        Let

        3 \log x=t

        \frac{3}{x} d x=d t \Rightarrow \frac{d x}{x}=\frac{d t}{3}

Put in (1)

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

                                      =\frac{1}{3} \sin ^{-1}\left(\frac{t}{2}\right)+c \quad\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]

                                       =\frac{1}{3} \sin ^{-1}\left(\frac{3 \log x}{2}\right)+c

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