Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 24 maths textbook solution.

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 24 maths textbook solution.

Answers (1)

Answer:- \frac{3\pi}{8}

Hints:-  You must know the integration rules of trigonometric functions and its limits.

Given:-  \int_{-\pi / 2}^{\pi / 2} \sin ^{4} x \cdot d x

Solution : \int_{-\pi / 2}^{\pi / 2} \sin ^{4} x \cdot d x

    \begin{aligned} &=\int_{-\pi / 2}^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\cos ^{2} 2 x-2 \cos ^{2} 2 x\right) \cdot d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\frac{1+\cos ^{4} x}{2}-2 \cos 2 x\right) \cdot d x \end{aligned}

    \begin{aligned} &=\frac{3}{8} \int_{-\pi / 2}^{\pi / 2} d x+\frac{1}{8} \int \cos ^{4} x \cdot d x-\frac{1}{2} \int \cos 2 x \cdot d x \\ &\left.\left.\left.=\frac{3}{8} x .\right]_{-\pi / 2}^{\pi / 2}+\frac{1}{32} \sin ^{4} x\right]_{-\pi / 2}^{\pi / 2}-\frac{1}{2} \sin 2 x\right]_{-\pi / 2}^{\pi / 2} \\ &=\frac{3 \pi}{8} \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 Board Exam Date 2027 Highlights: Class 10, 12 exam schedule soon at cbse.gov.in
anam-khan

CBSE Board Exams 2027: LOC submission for Classes 10, 12 begins tomorrow; check dates and fees
anam-khan

CBSE Board Exam Schedule 2027 Update : Class 10, 12 exam dates soon; LOC submission begins

Latest Question