Arctangent over sine, how do you even integrate this??

Put $2x=t$ to transform the integral to $\displaystyle \dfrac{1}{6} \int_{0}^{\pi/2} \dfrac{\arctan(5 \sin t)}{\sin t } dt$

Define $I(n) = \displaystyle \int_{0}^{\pi/2} \dfrac{\arctan (n \sin t)}{\sin t} dt$

Derivating with respect to $n$ ,

$I'(n) = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{1+n^2 \sin^2t} dt$

= $\frac{1}{n^2+1} \displaystyle \int_{0}^{\pi/2} \dfrac{\sec^2 t dt}{\tan^2 t + \frac{1}{n^2+1}}$

Put $\tan t = z$ to get, and use $\displaystyle \int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a} \arctan(x/a)$ to get

$I'(n) = \dfrac{\pi}{2 \sqrt{n^2+1}}$

Now, integrate both sides with respect to $n$ to get

$I(n) = \frac{\pi}{2} \ln(n+\sqrt{n^2+1}) + c$

As $I(0) = \displaystyle \int_{0}^{\pi/2} \dfrac{\arctan (0 \sin t)}{\sin t} dt = \int_{0}^{\pi/2} 0 dt = 0$ , we get $c = 0$ .

Hence, $I(n) = \frac{\pi}{2} \ln \bigg(n+\sqrt{n^2+1}\bigg)$

Replace $n=5$ and divide by 6 to get the answer.