Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 4 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 4 maths

Answers (1)

Answer:

4
Given:

\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x

Hint:

Using \int \sin x \; d x, \int \cos x \; d x

Explanation:  

Let

I=\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x

\begin{aligned} &{\left[\sin 2 \theta=2 \sin \theta \cos \theta, \quad \sin ^{2} \theta+\cos ^{2} \theta=1\right]} \\\\ &=\int_{0}^{2 \pi} \sqrt{\left(\sin {x} / 4+\cos {x} / 4\right)^{2}} d x \\\\ &=\int_{0}^{2 \pi}\left(\sin x / 4+\cos {x} / 4\right) d x \end{aligned}

\begin{aligned} &=\int_{0}^{2 \pi} \sin x / 4 d x+\int_{0}^{2 \pi} \cos ^{x} / 4 d x \\\\ &=\left[-4 \cos \frac{x}{4}\right]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \end{aligned}

\begin{aligned} &=-4 \cos [x / 4]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \\\\ &=\left[-4 \cos \frac{\pi}{2}+4 \cos 0\right]+\left[4 \sin \frac{\pi}{2}-4 \sin 0\right] \end{aligned}

\begin{aligned} &=0+4+0 \\\\ &=4 \end{aligned}

 

Posted by

infoexpert26

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