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 Fill in the blanks question 11

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 11

Answers (1)

Answer: \frac{\pi }{8}

Hint: Use \int \frac{1}{1+x^{2}} d x

Given: \int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x

Solution:  

\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x

Put

    \begin{aligned} &\sin ^{2} x=t \\\\ &2 \sin x \cos x \; d x=d t \\\\ &\sin x \cos x \; d x=\frac{d t}{2} \end{aligned}

When \mathrm{x}=0, \mathrm{t}=0

When \mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1

I=\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x

    \begin{aligned} &=\int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{d t}{2} \\\\ &=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1} \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\\\ &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}\left(\tan \frac{\pi}{4}\right)-\tan ^{-1}(\tan 0)\right] \\\\ &=\frac{1}{2}\left(\frac{\pi}{4}\right)-\frac{1}{2}(0) \\\\ &=\frac{\pi}{8} \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

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