Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 43.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 43.

Answers (1)

Answer  : \frac{3}{4}

Hint   : use indefinite formula and the limit to solve this integral

Given : \int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta

Solution : I=\int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta

\begin{aligned} &I==\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos ^{3} 2 \theta} \cdot \sin 2 \theta d \theta \\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sin 2 \theta}{\cos 2 \theta} \cdot \frac{1}{\cos ^{2} 2 \theta} d \theta \\ &I=\int_{0}^{\frac{\pi}{6}} \tan 2 \theta \cdot \sec ^{2} 2 \theta d \theta \end{aligned}

put \tan 2\theta =t \Rightarrow \sec^22\theta .2d\theta = dt \Rightarrow \sec^22\theta d\theta = \frac{dt}2{}

when \theta =0 then t=0 and when \theta = \frac{\pi}{6} then t=3

Therefore ,

\begin{aligned} &I=\int_{0}^{\sqrt{3}} t \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{\sqrt{3}} t d t \\ &=\frac{1}{2}\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\sqrt{3}} \end{aligned}

\begin{aligned} &=\frac{1}{2} \cdot \frac{1}{2}\left[t^{2}\right]_{0}^{\sqrt{3}} \\ &=\frac{1}{4}\left[\sqrt{3}^{2}-0^{2}\right] \\ &I=\frac{1}{4} \times 3 \\ &I=\frac{3}{4} \end{aligned}

Posted by

infoexpert24

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 Supplementary Exam 2026: 'Chapter-wise weightage ignored,' allege students; seek grace marks
anam-khan

CBSE SARAS 7.0: Affiliation application deadline extended to September 30 for 2027-28 session
anam-khan

CBSE Class 12 supplementary exam 2026 on July 28; major subjects, timings, exam day guidelines

Latest Question