Given,
$$
I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda \tan ^{-1}\left(\frac{1}{2}\right)
$$
Now, $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}$
$$
\begin{aligned}
& =\int_0^{\pi / 2} \frac{d x}{5+3 \times \frac{2 \tan x / 2}{1+\tan ^2 x / 2}} \\
& =\int_0^{\pi / 2} \frac{\left.d \text { put } \sin x=\frac{2 \tan x / 2}{1+\tan ^2 x / 2}\right]}{\frac{5+5 \tan ^2 x / 2+6 \tan x / 2}{1+\tan ^2 x / 2}} \\
& =\int_0^{\pi / 2} \frac{\left(1+\tan ^2 x / 2\right) d x}{5+5 \tan ^2 x / 2+6 \tan x / 2} \\
\Rightarrow \quad I & =\int_0^{\pi / 2} \frac{\sec ^2 x / 2 d x}{5+5 \tan ^2 x / 2+6 \tan x / 2}
\end{aligned}
$$
On putting $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t$
When $x=\frac{\pi}{2} \Rightarrow t=1$ and $x=0 \Rightarrow t=0$,
we get
$I=\int_0^1 \frac{2 d t}{5+5 t^2+6 t}$


$$
\begin{aligned}
& =\frac{2}{5} \int_0^1 \frac{d t}{1+t^2+\frac{6}{5} t} \\
& =\frac{2}{5} \int_0^1 \frac{d t}{\left(t^2+2 \cdot\left(\frac{3}{5}\right) t+\frac{9}{25}\right)+1-\frac{9}{25}} \\
& =\frac{2}{5} \int_0^1 \frac{d t}{\left(t+\frac{3}{5}\right)^2+\frac{16}{25}} \\
& =\frac{2}{5} \int_0^1 \frac{d t}{\left(t+\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2} \\
& =\left|\frac{2}{5} \cdot \frac{5}{4} \tan ^{-1} \frac{t+\frac{3}{5}}{\frac{4}{5}}\right|_0^1=\left|\frac{1}{2} \tan ^{-1}\left(\frac{5 t+3}{4}\right)\right|_0^1 \\
& =\frac{1}{2}\left[\tan ^{-1}\left(\frac{5+3}{4}\right)-\tan ^{-1} \frac{3}{4}\right] \\
& =\frac{1}{2}\left[\tan ^{-1} \frac{8}{4}-\tan ^{-1} \frac{3}{4}\right] \\
& =\frac{1}{2}\left[\tan ^{-1} 2-\tan ^{-1} \frac{3}{4}\right] \\
&
\end{aligned}
$$


$$
\begin{aligned}
& =\frac{1}{2} \tan ^{-1}\left(\frac{2-\frac{3}{4}}{1+2 \times \frac{3}{4}}\right) \\
& =\frac{1}{2} \tan ^{-1}\left(\frac{8-3 / 4}{\frac{2+3}{2}}\right) \\
& =\frac{1}{2} \tan ^{-1}\left(\frac{\frac{5}{5}}{\frac{5}{2}}\right)=\frac{1}{2} \tan ^{-1}\left(\frac{1}{2}\right) \\
\therefore \quad \lambda & =\frac{1}{2}
\end{aligned}
$$