Question

Area (in $sq.$ unit) of region bounded by $y=2\cos x,y=3\tan x$ and $y-$axis is

• A. $1+3ln\left(\frac{2}{\sqrt{3}}\right)$
• B. $1+\frac{3}{2}ln3-3ln2$
• C. $1+\frac{3}{2}ln3-ln2$
• D. $ln\left(\frac{3}{2}\right)$

Answer 1

Answer: (B)

$1+\frac{3}{2}ln3-3ln2$

Given that,

$y=2\cos x,y=3\tan x$

$2\cos x=3\tan x$

$2\cos x=3\frac{\sin x}{\cos x}$

$2{\cos }^{2}x=3\sin x$

$2(1-{\sin }^{2}x)=3\sin x$

$2-2{\sin }^{2}x=3\sin x$

$2{\sin }^{2}x+3\sin x-2=0$

On solving,

$x=\frac{\pi }{6}$ and y-axis

So, $x=0$

then,

${\int }_0^{\pi /6}(2\cos x-3\tan x)dx$

$=2{\int }_0^{\pi /6}\cos xdx-3{\int }_0^{\pi /6}\tan xdx$

$=2[\sin x{]}_0^{\pi /6}-3[log\sec x{]}_0^{\pi /6}$

$=2[\sin \frac{\pi }{6}-\sin \theta ]-3[log\sec \frac{\pi }{6}-log\sec 0]$

$=2[\frac{1}{2}-0]-3[log\frac{2}{\sqrt{3}}-log1]$

$=\frac{2}{2}-0-3[log2-log\sqrt{3}-0]$

$=1-3[log2-log\sqrt{3}]$

$=1-3log2+3log\sqrt{3}$

$=1-3log2+3log(3{)}^{\frac{1}{2}}$

$=1-3log2+\frac{3}{2}log3$

$=1+\frac{3}{2}log3-3log2$.