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 R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 9 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 9 Maths Textbook Solution.

Answers (1)

Answer: \frac{-1}{2} x \cos 2 x+\frac{1}{4} \sin 2 x+c

Hint: Use integrate on by parts

         \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x

Given: I=\int x \sin 2 x d x

Solution: I=\int x \sin 2 x d x

                \Rightarrow d(\cos 2 x)=-2 \sin 2 x d x

We can integrate by parts in this way

               \begin{aligned} &=\int x \sin 2 x d x \\ &=\frac{-1}{2} \int x d(\cos 2 x) \\ &=\frac{-1}{2} x \cos 2 x+\frac{1}{2} \int \cos 2 x d x \\ &=\frac{-1}{2} x \cos 2 x+\frac{1}{4} \sin 2 x+c \end{aligned}

 

Posted by

infoexpert21

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