Answer
\[\frac{1}{2}\sin 2x+C\]
Work Step by Step
Let \[I=\int\frac{1-\tan^2 x}{\sec^2 x}dx\]
\[I=\int\frac{1-\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}}dx\]
\[I=\int\left[\cos^2 x-\sin^2 x\right]dx\]
\[\left[\cos^2 x-\sin^2 x=\cos 2x\right]\]
\[I=\int\cos 2x\:dx=\frac{1}{2}\sin 2x+C\]
Hence $I=\frac{1}{2}\sin 2x+C$