Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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}

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question