$\int \frac{1-\tan^2 x}{1+\tan^2 x}dx$?1$\frac{\sin 2x}{2}+C$2$\frac{\cos 2x}{2}+C$3$\frac{\sin 2x}{4}+C$4$\sin 2x+C$Answer & explanation+Correct answer: option 1$\int \frac{1-\tan^2 x}{1+\tan^2 x}dx$=$\int \frac{1-\tan^2 x}{\sec^2 x}dx=\int (\cos^2x-\sin^2x)dx=\int \cos 2xdx=\frac{sin 2x}{2}+C$
