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 19 Definite Integrals Exercise 19.4 (b) Question 17 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 17 textbook solution.

Answers (1)

Answer:- \frac{\pi^{2}}{4}

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

Given:-  \int_{0}^{\pi} x \cos ^{2} x d x

Solution :  \int_{0}^{\pi} x \cos ^{2} x d x=I

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \cos ^{2}(\pi-x) d x \\ &I=\int_{0}^{\pi}(\pi-x) \cos ^{2} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \cos ^{2} x \cdot d x-\int_{0}^{\pi} x \cos ^{2} x \cdot d x \end{aligned}

\begin{aligned} &I=\pi \int_{0}^{\pi} \cos ^{2} x \cdot d x-I \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x+1}{2} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x}{2} \cdot d x+\int_{0}^{\pi} \frac{1}{2} \cdot d x \end{aligned}

\begin{gathered} =\pi\left[\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\pi}+\frac{1}{2}[x]_{0}^{\pi}\right] \\ 2 I=\frac{\pi}{4}[\sin 2 \pi-\sin 0]+\frac{\pi}{2}[\pi] \end{gathered}

\begin{aligned} &2 I=\frac{\pi^{2}}{4} \\ &I=\frac{\pi^{2}}{4} \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 Compartment Result 2026 LIVE: 12th supplementary marksheet out at results.digilocker.gov.in
anam-khan

CBSE Class 12 supplementary results 2026 out; 53.08% students pass
anam-khan

CBSE Class 12 Supplementary Exam 2026: 'Chapter-wise weightage ignored,' allege students; seek grace marks

Latest Question