Q116 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
$\int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$
- A$\frac{1}{2} \log 3$
- Correct answerB$\frac{1}{3} \log 2$
- C$2 \log 3$
- D$\frac{1}{2} \log \frac{3}{2}$
Solution
We have, $\int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$
$=\int_0^{\pi / 2} \frac{d x}{4+5\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)}$
$=\int_0^{\pi / 2} \frac{\left(1+\tan ^2 \frac{x}{2}\right)}{4+4 \tan ^2 \frac{x}{2}+10 \tan \frac{x}{2}} d x$
$=\frac{1}{4} \int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2}}{\tan ^2 \frac{x}{2}+\frac{5}{2} \tan \frac{x}{2}+1} d x$
let $\tan \frac{x}{2}=t$
$\Rightarrow \quad \sec ^2 \frac{x}{2}\left(\frac{1}{2}\right) d x=d t$
$\Rightarrow \sec ^2 \frac{x}{2} d x=2 d t$ at $x=0, t=0$
and $\quad x=\frac{\pi}{2}, t=1=\frac{2}{4} \int_0^1-\frac{1}{t^2+\frac{5}{2} t+1} d t$
$=\frac{1}{2} \int_0^1 \frac{d t}{t^2+\frac{5}{2} t+1+\frac{25}{16}-\frac{25}{16}}$
$=\frac{1}{2} \int_0^1 \frac{d t}{\left(t+\frac{5}{4}\right)^2-\frac{9}{16}}=\frac{1}{2} \int_0^1 \frac{d t}{\left(t+\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2}$
$=\frac{1}{2}\left[\frac{1}{2 \times\left(\frac{3}{4}\right)} \log \left|\frac{t+\frac{5}{4}-\frac{3}{4}}{t+\frac{5}{4}+\frac{3}{4}}\right|\right]_0^1$
$=\frac{1}{3}\left[\log \left|\frac{t+\frac{1}{2}}{t+2}\right|\right]_0^1$
$=\frac{1}{3}\left[\log \left(\frac{3 / 2}{3}\right)-\log \left(\frac{1 / 2}{2}\right)\right]$
$=\frac{1}{3}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{4}\right)\right]=\frac{1}{3}\left[\log \left(\frac{\frac{1}{2}}{\frac{1}{4}}\right)\right]$
$=\frac{1}{3} \log 2$
Hence, the correct answer is option: 'B'.