Evaluate $\int\limits_{0}^{\pi/4} \frac{x dx}{1 + \cos 2x + \sin 2x}$.

$\frac{\pi}{16} \log 2$

The correct answer is Option (4) → $\frac{\pi}{16} \log 2$

$I = \int\limits_{0}^{\frac{\pi}{4}} \frac{x \, dx}{1 + \cos 2x + \sin 2x} \,\, dots(i)$

$\left[ \text{Using } \int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a - x) dx \right]$

$I = \int\limits_{0}^{\frac{\pi}{4}} \frac{\left( \frac{\pi}{4} - x \right) dx}{1 + \cos\left( \frac{\pi}{2} - 2x \right) + \sin\left( \frac{\pi}{2} - 2x \right)}$

$I = \int\limits_{0}^{\frac{\pi}{4}} \frac{\left( \frac{\pi}{4} - x \right) dx}{1 + \sin 2x + \cos 2x} \,\,\, \dots(ii)$

Now add eqs. (i) and (ii)

$2I = \int\limits_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{1 + \sin 2x + \cos 2x} dx$

$2I = \frac{\pi}{4} \int\limits_{0}^{\frac{\pi}{4}} \frac{dx}{2\cos^2 x + 2\sin x \cos x}$

$[∵1 + \cos 2\theta = 2\cos^2 \theta, \sin 2\theta = 2\sin \theta \cos \theta]$

$\Rightarrow 2I = \frac{\pi}{8} \int\limits_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\cos^2 x}}{1 + \tan x}$

$\Rightarrow I = \frac{\pi}{16} \int\limits_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{1 + \tan x} dx$

Let $1 + \tan x = t$

$\sec^2 x dx = dt$

New limit:

$1 + \tan 0 = t \Rightarrow t = 1$

$1 + \tan \frac{\pi}{4} = t \Rightarrow t = 2$

$\Rightarrow I = \frac{\pi}{16} \int\limits_{1}^{2} \frac{dt}{t} \Rightarrow I = \frac{\pi}{16} [\log |t|]_1^2$

$I = \frac{\pi}{16} \log(2) - \log(1)$

$I = \frac{\pi}{16} \log 2$