Paneer Definite Masala 9

Let, $\large I = \int_0^{\pi /4} \arctan \left( \dfrac{2\cos^2 x}{2-\sin2x} \right) \sec^2 x \, dx$

Subtitute $\tan x = t$ , differentiating, $\sec^2 x \,dx =dt$ . So, $\begin{aligned}I &= \int_0^1 \arctan\left(\dfrac{2}{(1+t^2)(2-\dfrac{2t}{1+t^2})}\right)dt\\ &= \int_0^1 \arctan\left(\dfrac{1}{1+t^2-t}\right)dt\\ &= \int_0^1 \arctan\left(\dfrac{t+(1-t)}{1-t(1-t)}\right)dt\end{aligned}$

Now now, $\arctan x + \arctan y = \begin{cases}\arctan \left(\dfrac{x+y}{1-xy}\right) & \text{ for } &xy<1\\\pi+\arctan \left(\dfrac{x+y}{1-xy}\right) & \text{ for }& xy>1\end{cases}$

But since $t(t-1) < 1$ for all real values for $t$ , we can use the first case.

So, $\begin{aligned}I &= \int_0^1 \arctan t + \arctan (1-t)\, dt\\ &=2\int_0^1 \arctan t\,dt\\ &=2\left(t\,\arctan t\right)_0^1 - \int_0^1 \dfrac{2t}{1+t^2}\, dt\\ &=\Large\boxed{\dfrac\pi2 - \ln 2}\end{aligned}$