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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question