$\int\limits_{0}^{4π}\frac{dx}{\cos^2x(2+\tan^2x)}$ is equal to:
Answer & explanation
Correct answer: option 2
$\int\limits_{0}^{4π}\frac{dx}{\cos^2x(2+\tan^2x)}=\int\limits_{0}^{4π}\frac{\sec^2x}{2+\tan^2x}dx=8\int\limits_{0}^{\frac{π}{2}}\frac{\sec^2x}{2+\tan^2x}dx=8\int\limits_{0}^{\frac{π}{2}}\frac{dt}{2+t^2}=4\sqrt{2}\tan^{-1}\frac{t}{\sqrt{2}}|_0^∞=2\sqrt{2}π$