Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 5 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 5 maths textbook solution

Answers (1)

Answer: 1

Hint: Using \int \sin x \; \mathrm{dx}

Given:\int_{1 / \pi}^{2 /\pi } \frac{\sin (1 / x)}{x^{2}}dx

Solution:  

I=\int_{1 / \pi}^{2 /\pi } \frac{\sin (1 / x)}{x^{2}}dx

Put  \frac{1}{x}=t

        \begin{aligned} &\frac{-1}{x^{2}} d x=d t \\\\ &\frac{d x}{x^{2}}=-d t \end{aligned}

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

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

I=\int_{\pi}^{\pi / 2} \sin t(-d t)

    \begin{gathered} =-\int_{\pi}^{\pi / 2} \sin t \; dt\\\\ =-[-\cos t]_{\pi}^{\pi / 2} \end{gathered}

    \begin{aligned} &=-\left[-\cos \frac{\pi}{2}-(-\cos \pi)\right] \\\\ &=\cos \frac{\pi}{2}-(\cos \pi) \\\\ &=0-(-1) \\\\ &=1 \end{aligned}            [\because \cos \pi=-1]

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

‘Feeling Devastated’: CBSE 12th students allege low marks due to on-screen marking; call for manual evaluation
anam-khan

'No separate prep for CBSE and JEE': Gurugram CBSE topper balances both with one routine, here's how
anam-khan

What does RT mean in CBSE result?

Latest Question