Question

The integral
$\int \sqrt{1+2\cot x\left(\cos ecx+\mathrm{cotx}\right)}dx\left(0\prec x\prec \frac{\pi }{2}\right)$ is equal to:

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

Answer 1

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

$y=\sqrt{1+\frac{2\cos x}{\sin x}\left(\frac{1+\cos x}{\sin x}\right)}=\sqrt{1+\frac{2\cos x}{\sin x}.\frac{2{\cos }^{2}x/2}{\sin x}}$

$=\sqrt{1+\frac{2\cos x}{4{\sin }^2\frac{x}{2}{\cos }^2\frac{x}{2}}.2{\cos }^{2}x/2}=\sqrt{1+\frac{\cos x}{{\sin }^{2}x/2}}$

$=\frac{1}{\sin \frac{x}{2}}\left[\sqrt{1-2{\sin }^2\frac{x}{2}.+{\sin }^2\frac{x}{2}}\right]=\frac{1}{\sin \frac{x}{2}}\left[\sqrt{1-{\sin }^2\frac{x}{2}}\right]$

$=\cot x/2$

$\therefore I=\int \cot \left(x/2\right)dx=2\log \left(\sin \frac{x}{2}\right)+c$