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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 61 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 61 Maths Textbook Solution.

Answers (1)

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

Hint:Put \sin ^{-1}x=u

Given:\int \frac{x^{2}\sin ^{-1}x}{\left ( 1-x^{2} \right )^{\frac{3}{2}}}dx

Solution: Put

          \sin ^{-1}x=u\Rightarrow \frac{1}{\sqrt{1-x^{2}}}dx=du

          and

          x=\sin u

         \begin{aligned} &\therefore \int \frac{x^{2} \sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} d x=\int \frac{x^{2} \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}} d x \\ &\Rightarrow \int\left(\frac{u \sin ^{2} u}{1-\sin ^{2} u}\right) d u \\ &=\int \frac{u \sin ^{2} u}{\cos ^{2} u} d u=\int u \tan ^{2} u d u=\int u\left(\sec ^{2} u-1\right) d u \end{aligned}

           \begin{aligned} &=\int u \sec ^{2} u d u-\int u d u \\ &=u \times \int \sec ^{2} u d u-\int \frac{d}{d u} u \int \sec ^{2} u d u d u-\frac{u^{2}}{2}+c \\ &=u \tan u-\int 1 \times \tan u d u-\frac{u^{2}}{2}+c \\ &=u \tan u-\int \frac{\sin u}{\cos u} d u-\frac{u^{2}}{2}+c \end{aligned}

Put       \cos u=v\Rightarrow -\sin udu=dv\Rightarrow \sin udu=-dv

             \begin{aligned} &=u \tan u+\int \frac{d v}{v}-\frac{u^{2}}{2}+c \\ &=u \tan u+\ln v-\frac{u^{2}}{2}+c \\ &=u \frac{\sin u}{\cos u}+\ln \left(\sqrt{1-\sin ^{2} u}\right)-\frac{u^{2}}{2}+c \\ &=u \frac{\sin u}{\sqrt{1-\sin ^{2} u}}+\ln \left(\sqrt{1-\sin ^{2} u}\right)-\frac{u^{2}}{2}+c \\ &=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}+\log \left(\sqrt{1-x^{2}}\right)-\frac{\left(\sin ^{-1} x\right)^{2}}{2}+c \end{aligned}

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