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  • Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18. 8 question 40 maths

Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 40 maths

Answers (1)

Answer:

        \frac{1}{3}log\left | 2+3sin^{-1}x \right |+C

Hint:

        \frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}

Given:

        \int \! \frac{1}{\sqrt{1-x^{2}}(2+3sin^{-1}x)}dx                    ......(1)

Explanation:

Let

        2+3sin^{-1}x=t

        \frac{3}{\sqrt{1-x^{2}}}dx=dt

        \frac{dx}{\sqrt{1-x^{2}}}=\frac{dt}{3}

Put in (1) we get

        \frac{1}{3}\int \! \frac{dt}{t}=\frac{1}{3}log\left | t \right |+C

                      =\frac{1}{3}log\left | 2+3sin^{-1}x \right |+C

Posted by

Gurleen Kaur

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