Question

$\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{4+5\sin x}=$

• A. $\frac{1}{2}\log 3$
• B. $\frac{1}{3}\log 2$
• C. $2\log 3$
• D. $\frac{1}{2}\log \frac{3}{2}$

Answer 1

Answer: (B)

We have, $\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{4+5\sin x}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{4+5\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$
$=\underset{0}{\overset{\pi 2}{\int }}\frac{\left(1+{\tan }^2\frac{x}{2}\right)}{4+4{\tan }^2\frac{x}{2}+10\tan \frac{x}{2}}dx$
$=\frac{1}{4}\underset{0}{\overset{\pi 2}{\int }}\frac{\sec \frac{2x}{2}}{{\tan }^2\frac{x}{2}+\frac{5}{2}\tan \frac{x}{2}+1}dx$
let $\tan \frac{x}{2}=t$
$\Rightarrow {\sec }^2\frac{x}{2}\left(\frac{1}{2}\right)dx=dt$
$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$ at $x=0,t=0$ and
$x=\frac{\pi }{2},t=1=\frac{2}{4}\underset{0}{\overset{1}{\int }}-\frac{1}{t^2+\frac{5}{2}t+1}dt$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+\frac{5}{2}t+1+\frac{25}{16}-\frac{25}{16}}$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{{\left(t+\frac{5}{4}\right)}^2-\frac{9}{16}}=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{{\left(t+\frac{5}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2}$
$=\frac{1}{3}{\left[\log \left|\frac{t+\frac{1}{2}}{t+2}\right|\right]}_0^1$
$=\frac{1}{3}\left[\log \left(\frac{3/2}{3}\right)-\log \left(\frac{1/2}{2}\right)\right]$
$=\frac{1}{3}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{4}\right)\right]$
$=\frac{1}{3}\left[\log \left(\frac{\frac{1}{2}4}{1}\right)\right]$
$=\frac{1}{3}\log 2$