Question

The value of $\underset{\frac{π}{3}}{\overset{\frac{π}{2}}{∫}}\frac{(2+3\sin x)}{\sin x(1+\cos x)}dx$ is equal to

• A. $\frac{7}{2}−\sqrt{3}−{\log }_e\sqrt{3}$
• B. $\frac{10}{3}−\sqrt{3}−{\log }_e\sqrt{3}$
• C. $\frac{10}{3}−\sqrt{3}+{\log }_e\sqrt{3}$
• D. $−2+3\sqrt{3}+{\log }_e\sqrt{3}$

Answer 1

Answer: (C)

$\underset{π/3}{\overset{π/2}{∫}}\left(\frac{2+3\sin x}{\sin x(1+\cos x)}\right)dx=2\underset{π/3}{\overset{π/2}{∫}}\frac{dx}{\sin x+\sin x\cos x}+3$
$3\underset{π/3}{\overset{π/2}{∫}}\frac{dx}{1+\cos x}$
$\underset{π/3}{\overset{π/2}{∫}}\frac{dx}{1+\cos x}=\underset{π/3}{\overset{π/2}{∫}}\frac{1−\cos x}{{\sin }^{2}x}dx$
$=\underset{π/3}{\overset{π/2}{∫}}\left({\text{cosec}}^{2}x−\cot x\text{cosec}x\right)dx$
$=(\mathrm{cosec}x−\cot x)\underset{π/3}{\overset{π/2}{∫}}=(1)−\left(\frac{2}{\sqrt{3}}−\frac{1}{\sqrt{3}}\right)=1−\frac{1}{\sqrt{3}}$
$\underset{π/3}{\overset{π/2}{∫}}\frac{dx}{\sin x(1+\cos x)}=$
$∫\frac{dx}{(2\tan x/2)\left(1+1−{\tan }^{2}x/2\right)}$
$=∫\frac{\left(1+{\tan }^{2}x/2\right){\sec }^{2}x/2dx}{2\tan x/2}$
$\tan x/2=t$
$\sec x/2\frac{1}{2}dx=dt$
$\frac{1}{2}∫\left(\frac{1+t^2}{t}\right)dt=\frac{1}{2}{\left[\ln +{\frac{t}{2}}^2\right]}_{\frac{1}{\sqrt{3}}}^1$
$=\frac{1}{2}\left[\left(0+\frac{1}{2}\right)−\left(\ln \frac{1}{\sqrt{3}}+\frac{1}{6}\right)\right]=\left(\frac{1}{3}+\ln \sqrt{3}\right)\frac{1}{2}$
$=\left(\frac{1}{6}+\frac{1}{2}\ln \sqrt{3}\right)$
$2\left(\frac{1}{6}+\frac{1}{2}\ln \sqrt{3}\right)+3\left(1−\frac{1}{\sqrt{3}}\right)$
$=\frac{1}{3}+\ln \sqrt{3}+3−\sqrt{3}=\frac{10}{3}+\ln \sqrt{3}−\sqrt{3}$