Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 42 maths textbook solution

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 42 maths textbook solution.

Answers (1)

Answer:-  \frac{2}{\pi}+\frac{1}{\pi^{2}}

Hints:-  You must know the integration rules of trignometric functions.

Given:-  \int_{0}^{3 / 2}|x \sin \pi x| \cdot d x

Solution : \int_{0}^{3 / 2}|x \sin \pi x| \cdot d x

               \begin{aligned} &=\int_{0}^{1} x \sin \pi x \cdot d x-\int_{1}^{3 / 2} x \sin \pi x \cdot d x \\ &=\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{0}^{1}-\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi}\right]_{1}^{3 / 2} \end{aligned}

                \begin{aligned} &=\left[\frac{-1 \cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}-0+0\right]-\left[\frac{-3}{2} \frac{\cos ^{3 \pi / 2}}{\pi}+\frac{\cos \pi}{\pi}+\frac{\sin ^{3 \pi / 2}}{\pi^{2}}-\frac{\sin \pi}{\pi}\right]_{1}^{3 / 2} \\ &=\frac{1}{\pi}+\frac{3}{2} \cdot 0+\frac{1}{\pi}+\frac{1}{\pi^{2}}+0 \\ &=\frac{2}{\pi}+\frac{1}{\pi^{2}} \end{aligned}

Posted by

infoexpert23

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

CBSE Class 12 Board Exam 2026 LIVE: 12th economics paper ends; pattern, timings, answer key updates
anam-khan

CBSE Results 2026: Board warns teachers of legal action for sharing 'misleading' information on evaluation
anam-khan

CBSE 12th Hindi Analysis 2026: Paper ‘easy, balanced’ with direct questions, say experts

Latest Question