Answer
$$\int\frac{1-\tan^2x}{\sec^2x}dx=\frac{1}{2}\sin2x+C$$
Work Step by Step
$$A=\int\frac{1-\tan^2x}{\sec^2x}dx$$ $$A=\int\frac{1-\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}}dx$$ $$A=\int\frac{\frac{\cos^2 x-\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}}dx$$ $$A=\int\frac{(\cos^2 x-\sin^2 x)\cos^2 x}{\cos^2 x}dx$$ $$A=\int(\cos^2 x-\sin^2 x)dx$$ $$A=\int\cos2x dx$$ (for $\cos2x=\cos^2 x-\sin^2 x$) $$A=\frac{1}{2}\int\cos2x d(2x)$$ $$A=\frac{1}{2}\sin2x+C$$
