• # question_answer $\int\limits_{\pi /4}^{\pi /2}{\sqrt{2+\sqrt{2+2\cos 4x}}}\,\,dx$ is equal to: A) $\sqrt{2}$ B) $\sqrt{2}\,(\sqrt{2}-1)$ C) 2 D) none of these

 $l=\int\limits_{\pi /4}^{\pi /2}{\sqrt{2+\sqrt{2+2\cos 4x}}}\,\,dx=\int\limits_{\pi /4}^{\pi /2}{\sqrt{2-2\cos 2x}\,\,dx}$ $=2\int\limits_{\pi /4}^{\pi /2}{\sin x\,\,dx}$      $=2\,(-\cos x)_{\pi /4}^{\pi /2}=\sqrt{2}$