# Get Answers to all your Questions

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

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}