Question

The integral $\int \sqrt{1+2\cot x(cosecx+\cot x)}dx(0<x<\frac{\pi }{2})$ is equal to?

• A. $4log\left(\sin \frac{x}{2}\right)+c$
• B. $2log\left(\sin \frac{x}{2}\right)+c$
• C. $2log\left(\cos \frac{x}{2}\right)+c$
• D. $4log\left(\cos \frac{x}{2}\right)+c$

Answer 1

The correct option is B $2log\left(\sin \frac{x}{2}\right)+c$

We have,

$I=\int \sqrt{1+2cotx(cosecx+cotx)}dx$

We know that

$cose{c}^{2}x-co{t}^{2}x=1$

then,

$I=\int \sqrt{cose{c}^{2}x-co{t}^{2}x+2cotx(cosecx+tanx)}dx$

$I=\int \sqrt{(cose{c}^{2}x-co{t}^{2}x+2cotxcosecx+2co{t}^{2}x)}dx$

$I=\int (cose{c}^{2}x+co{t}^{2}x+2cotxcosecx{)}^{\frac{1}{2}}dx$

$I=\int \sqrt{(cosecx+cotx{)}^2}dx$

$\because (a+b{)}^2=a^2+b^2+2ab$

$I=\int (cosecx+cotx)dx$

$I=\int (\frac{1}{sinx}+\frac{cosx}{sinx})dx$

$=\int \left(\frac{1+cosx}{sinx}\right)dx$

$=\int \frac{1+2co{s}^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}dx$

$\because cosx=2co{s}^2\frac{x}{2}-1$

$\because sinx=2sin\frac{x}{2}cos\frac{x}{2}$

$=\int \frac{2co{s}^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}dx$

$=\int \frac{cos\frac{x}{2}}{sin\frac{x}{2}}dx$

$=cot\frac{x}{2}dx$

on integrating and we get

$I=log\left|\frac{sin\frac{x}{2}}{\frac{1}{2}}\right|+C$

$\therefore \int cotxdx=logsinx$

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

Hence, this is the answer.