I am so brain dead on how to integrate sec(2x)dx. I converted it to the integration of 1/(cos^2(x)-sin^2(x)) and I am so blank on how to go on from here.

Question

anonymous · Answer

$$\int\sec 2x~dx$$
Let $u=2x$, so $\dfrac{1}{2}du=dx$.
$$\frac{1}{2}\int \sec u~du$$
The next step isn't incredibly intuitive, so it helps to recall your trigonometric derivatives:
$$\frac{1}{2}\int \sec u\cdot\frac{\sec u+\tan u}{\sec u+\tan u}~du$$
When you distribute the $\sec u$, your numerator will be $\sec^2u+\sec u\tan u$, which is the derivative of the denominator, and so you can use another substitution to proceed.