330 followers problem-2

$\begin{aligned} I(a) & = \int_0^\pi \frac x{1+a\sin^2 x}dx & \small {\color{#3D99F6}\text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\pi \frac x{1+a\sin^2 x} + \frac {\pi -x}{1+a \ {\color{#3D99F6}\sin^2(\pi - x)}} dx & \small {\color{#3D99F6}\text{Note that }\sin (\pi-x) = \sin x} \\ & = \frac 12 \int_0^\pi \frac \pi{1+a\sin^2 x} dx & \small {\color{#3D99F6}\text{As the integrand is symmetrical on } x = \frac \pi 2} \\ & = \pi \int_0^\frac \pi 2 \frac 1{1+a\sin^2 x} dx & \small {\color{#3D99F6}\text{Multiply up and down by } \sec^2 x} \\ & = \pi \int_0^\frac \pi 2 \frac {\sec^2 x}{\sec^2 x +a\tan^2 x} dx \\ & = \pi \int_0^\frac \pi 2 \frac {\sec^2 x}{1+(1 +a) \tan^2 x} dx & \small {\color{#3D99F6}\text{Let } t = \tan x, \ dt = \sec^2 x \ dx} \\ & = \pi \int_0^\infty \frac {dt}{1+(1 +a) t^2} & \small {\color{#3D99F6}\text{Let } u = \sqrt{1+a}\ t, \ du = \sqrt{1+a}\ dt} \\ & = \frac \pi{ \sqrt{1+a}} \int_0^\infty \frac {du}{1+u^2} \\ & = \frac {\pi \tan^{-1} u}{ \sqrt{1+a}} \bigg|_0^\infty \\ & = \frac {\pi^2}{2\sqrt{1+a}} \end{aligned}$

$\implies I(8) = \dfrac {\pi^2}{2\sqrt{1+8}} = \dfrac {\pi^2}6 = \zeta (2) \implies k = \boxed{2}$

With some basic applications it can be derived that $\displaystyle I_a =\pi\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+a\sin^2 x} =\pi\int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x dx}{1+\tan^2 x(1+a)} = \pi\int_{0}^{\infty}\frac{dt}{1+t^2(1+a)} = \frac{\pi^2}{2\sqrt{1+a}}$

Thus , $I_8=\zeta(2)$