Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 12 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 12 maths

Answers (1)

Answer:

\frac{\pi }{12}

Given:

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x

Hint:

Using \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x


Explanation:  

Let

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x \\\\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\frac{\cos x}{\sin x}}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right] \end{aligned}

    =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots(i)

We have a property that

\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

Using this, we get

=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x

=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x,\left[\therefore \sin \left(\frac{\pi}{2}-x\right)=\cos x, \quad \cos \left(\frac{\pi}{2}-x\right)=\sin x\right]

I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i i)

Adding (i) and (ii)

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}

\begin{aligned} &I=\frac{1}{2}[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right] \end{aligned}

\begin{aligned} &I=\frac{1}{2}\left[\frac{2 \pi-2 \pi}{6}\right] \\\\ &I=\frac{1}{2} \times \frac{\pi}{6} \\\\ &=\frac{\pi}{12} \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

CBSE Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in
anam-khan

CBSE Board 2025: Class 10th, 12th supplementary date sheet out; exams from July 15
anam-khan

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee

Latest Question