Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 50 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 50  Maths Textbook Solution.

Answers (1)

Answer: \frac{\sin 3x}{18}-\frac{x\cos 3x}{6}+\frac{x\cos x}{2}-\frac{\sin x}{2}+c

Hint: \int sin\: xdx=\cos \: x&\int \cos \: xdx=-\sin \: x

Given: \int\: x \sin x\cos 2xdx

Solution:

            I=\int\: x \sin x\cos 2xdx

            I=\frac{1}{2}\int\: 2x \sin x\cos 2xdx

            =\frac{1}{2}\int\: x2 \sin x\cos 2xdx

            \begin{aligned} &=\frac{1}{2} \int x[\sin (x+2 x)+\sin (x-2 x)] d x \\ &=\frac{1}{2} \int x \sin 3 x d x-\frac{1}{2} \int x \sin x d x \\ &=\frac{1}{2}\left[x \frac{\cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x\right]-\frac{1}{2}\left[x \cos x+\int \cos x d x\right] \\ &I=\frac{1}{2}\left[-x \frac{\cos 3 x}{3}+\frac{\sin 3 x}{9}\right]-\frac{1}{2}[-x \cos x+\sin x] \\ &=\frac{1}{2}\left[\frac{\sin 3 x}{9}-\frac{x \cos 3 x}{3}+x \cos x-\sin x\right]+c \\ &=\frac{\sin 3 x}{18}-\frac{x \cos 3 x}{6}+\frac{x \cos x}{2}-\frac{\sin x}{2}+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