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 announces deadline for Class 10, 12 direct admissions, subject change
anam-khan

CBSE warns about unauthorised sources offering ‘quick’ ways of getting duplicate marksheets, certificates
anam-khan

CBSE unveils 8 major changes for Class 10, 12 board exams; what students need to know

Latest Question