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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 24.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 24.
 

Answers (1)

Answer: \frac{1}{2}-\frac{\sqrt{3}}{12}\pi

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: I =\int_{0}^{\frac{1}{2}}\frac{x \sin^{-1}x}{\sqrt{1-x^2}}dx

Solution:I =\int_{0}^{\frac{1}{2}}\frac{x \sin^{-1}x}{\sqrt{1-x^2}}dx

Applying integration by parts method we get

=\left[\sin ^{-1} x \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^{2}}} d x\right]_{0}^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}}\left(\frac{d}{d x} \sin ^{-1} x \int \frac{x}{\sqrt{1-x^{2}}} d x\right) d x

In \int \frac{x}{\sqrt{1-x^2}} dx  put 1-x^2=t^2

\begin{aligned} &\rightarrow-2 x d x=2 t d t \\ &\rightarrow x d x=-t d t \\ &\int \frac{x}{\sqrt{1-x^{2}}} d x=-\int \frac{1}{\sqrt{t^{2}}} t d t \\ &=-\int \frac{1}{t} \cdot t d t \\ &=-\int 1 d t \end{aligned}

=\int t^0 dt                                  \left [ \int x^n dx=\frac{x^n+1}{n+1}+c \right ]

=-\left [ \frac{t^0+1}{0+1} \right ]

=-t

\begin{aligned} &\int \frac{x}{\sqrt{1-x^{2}}} d x=-t=-\sqrt{1-x^{2}} \quad \ldots(i) \quad\left[t^{2}=1-x^{2}\right] \\ &\left.I=\left[\sin ^{-1} x \cdot-\sqrt{1-x^{2}}\right]_{0}^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}}\left(\frac{1}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right)\right) d x \quad \text { [using }(i)\right]\left[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right] \end{aligned}

\begin{aligned} &=-\left[\sqrt{1-x^{2}} \cdot \sin ^{-1} x\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}} 1 d x \\ &=-\left[\sqrt{1-\left(\frac{1}{2}\right)^{2}} \cdot \sin ^{-1} \frac{1}{2}-\sqrt{1-0^{2}} \cdot \sin ^{-1} 0\right]+[x]_{0}^{\frac{1}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int 1 d x=x\right] \end{aligned}

\begin{aligned} &=-\left[\sqrt{1-\frac{1}{4}} \cdot \sin ^{-1} \frac{1}{2}-0\right]+\left[\frac{1}{2}-0\right]\\ &\left[\sin ^{-1} 0=0\right]\\ &=-\left[\sqrt{\frac{4-1}{4}} \frac{\pi}{6}\right]+\left[\frac{1}{2}\right] \quad\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right] \end{aligned}

\begin{aligned} &=-\frac{\sqrt{3}}{2} \frac{\pi}{6}+\frac{1}{2} \\ &=\frac{1}{2}-\frac{\sqrt{3} \pi}{12} \end{aligned}

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