Mathematics

# The integral $\displaystyle\int \sqrt{1+2\cot x(cosec x+\cot x)}dx\left(0 < x < \dfrac{\pi}{2}\right)$ is equal to?

$2log\left(\sin \dfrac{x}{2}\right)+c$

##### SOLUTION
We have,
$\displaystyle I= \int \sqrt{1+2cot x(cosec x+ cot x)}dx$
We know that
$cosec ^{2}x- cot^{2}x= 1$
then,
$\displaystyle I= \int \sqrt{cosec^{2}x- cot^{2}x+2 cotx(cosec x+ tan x)}dx$
$\displaystyle I= \int \sqrt{(cosec^{2}x- cot^{2}x+2 cotx\,cosec x+2 cot^2 x)}dx$
$\displaystyle I= \int (cosec^{2}x+cot^{2}x+2 cotx\, cosec x)^{\frac{1}{2}}dx$
$\displaystyle I= \int \sqrt{(cosec x+ cot x)^{2}}dx$
$\because (a+b)^{2}= a^{2}+b^{2}+2ab$
$\displaystyle I= \int (cosec x+ cot x)dx$
$\displaystyle I= \int (\frac{1}{sin x}+\frac{cos x}{sin x})dx$
$\displaystyle =\int (\frac{1+cos x}{sin x})dx$
$\displaystyle =\int \frac{1+2cos^{2}\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}dx$
$\displaystyle \because cos x= 2 cos^{2}\frac{x}{2}-1$
$\displaystyle \because sin x = 2sin\frac{x}{2}cos\frac{x}{2}$
$\displaystyle =\int \frac{2cos^{2}\frac{x}{2}}{2 sin\frac{x}{2}cos\frac{x}{2}}dx$
$\displaystyle =\int \frac{cos \frac{x}{2}}{sin \frac{x}{2}}dx$
$\displaystyle = cot \frac{x}{2}dx$
on integrating and we get
$\displaystyle I= log \left | \frac{sin\frac{x}{2}}{\frac{1}{2}} \right |+C$
$\displaystyle \therefore \int cot\,x\,dx= log\, sin x$
$\displaystyle I= 2 log(sin\frac{x}{2})+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\int^{2\pi}_0|\sin x|dx=?$
• A. $2$
• B. $1$
• C. None of these
• D. $4$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$
• A. is equal to one
• B. is equal to $\displaystyle \frac{1}{2}$
• C. can not be evaluated
• D. is equal to zero

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int\frac{x+2}{2x^{2}-7x+3}dx=$
• A. $\displaystyle \log|\frac{x-3}{2x-1}|+c$
• B. $\dfrac{1}{2}\displaystyle \log|\frac{x-3}{2x-1}|+c$
• C. $\dfrac{1}{2}\log|\displaystyle \frac{x-3}{\sqrt{2x-1}}|+c$
• D. $\displaystyle \log|\frac{x-3}{\sqrt{2x-1}}|+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate the function    $\cfrac {1}{\sqrt {(2-x)^2+1}}$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$