Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The integral

$\small \int \sqrt{1 + 2cot x ( cosec x + cot x) dx}$

$\small \left ( 0< x< \frac{\pi }{2} \right )$ is equal to

(where C is a constant of integration)

Option 1)
$4\log \left ( \sin \frac{x}{2} \right ) + C$

Option 2)
$2\log \left ( \sin \frac{x}{2} \right ) + C$

Option 3)
$2\log \left ( \cos \frac{X}{2} \right ) +C$

Option 4)
$4\log \left ( \cos \frac{x}{2} \right ) +C$

Answers (1)

best_answer

As we learnt in

Integrals for Trigonometric functions -

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos x \right ) =sinx$

$\therefore \int sinxdx=-cosx+c$

-

$\int \sqrt{\left ( 1+2\cot x\ cosecx\ \ +2\cot ^{2}x \right )dx}$

$=\int \sqrt{\left ( 1+\cot ^{2}x \right )+2\cot x\: cosecx+\cot ^{2}x}\:dx$

$=\int \sqrt{\left ( cosec^{2}x+2\cot x\:cosecx+\cot ^{2}x \right )}dx$

$=\int \left ( cosecx+\cot x \right )dx$

$=ln\ tan\frac{x}{2}+ln\ sin x$

$=ln\frac{sin\frac{x}{2}}{cos\frac{x}{2}}\times 2sin\frac{x}{2}cos\frac{x}{2}$

$=ln2\sin ^2\frac{x}{2}+C$

$=2\ ln\frac{\sin x}{2}+C'$

Option 1)

$4\log \left ( \sin \frac{x}{2} \right ) + C$

Incorrect option

Option 2)

$2\log \left ( \sin \frac{x}{2} \right ) + C$

Correct option

Option 3)

$2\log \left ( \cos \frac{X}{2} \right ) +C$

Incorrect option

Option 4)

$4\log \left ( \cos \frac{x}{2} \right ) +C$

Incorrect option

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Andhra Pradesh cut-offs drop for CSE, ECE, other BTech programmes this year; closing rank

anam-khan

JEE Main 2025 Live: Exam city slip soon at jeemain.nta.nic.in; admit card date, cut-off updates

anam-khan

BTech Admissions 2025: GUJCET registration ends on December 31; JEE Main, AEEE, WBJEE, SRMJEEE exam date

Latest Question