# Ways to evaluate ∫secθdθ\int \sec \theta \, \mathrm d \theta

The standard approach for showing $$∫secθdθ=ln|secθ+tanθ|+C\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$$ is to multiply by $$secθ+tanθsecθ+tanθ\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$$ and then do a substitution with $$u=secθ+tanθu = \sec \theta + \tan \theta$$.

I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It’s not very intuitive, nor does it seem to have applicability to any integration problem other than $$∫cscθdθ\int \csc \theta \,\mathrm d \theta$$. Does anyone know of another way to evaluate $$∫secθdθ\int \sec \theta \, \mathrm d \theta$$?

$$∫secxdx=∫cosxcos2xdx=∫cosx1−sin2xdx=12∫(11−sinx+11+sinx)cosxdx\int \sec x \,dx = \int \frac{\cos x}{\cos^2 x} \,dx = \int \frac{\cos x}{1-\sin^2 x} \,dx = \frac{1}{2} \int \left( \frac{1}{1-\sin x} + \frac{1}{1+\sin x} \right) \cos x dx$$
$$=12log|1+sinx1−sinx|+C.= \frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} \right| + C.$$
$$log|tan(π4+x2)|\log \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right|$$