# Kai’s exercises :)

$\displaystyle \int \frac{1}{\cos \theta} d\theta$

$\displaystyle t = \tan \frac{\theta}{2}$

$\displaystyle dt = \frac{1}{\cos^2 \frac{\theta}{2}}\frac{d\theta}{2}$

$\displaystyle \int \frac{dt}{t} = \int \frac{d\theta}{2\cos^2 \frac{\theta}{2}}\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = \int \frac{d\theta}{2\cos \frac{\theta}{2}}\frac{1}{\sin \frac{\theta}{2}} = \int \frac{d\theta}{2\cos \frac{\theta}{2}\sin \frac{\theta}{2}}$

Wrong.

We need to express $1/\cos x$ as a function of $\sin x / \cos x$ and $1 / \cos^2 x$.

$\displaystyle \tan^2 x + 1 = \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$.

Now we do:

$\displaystyle t = \tan \frac{\theta}{2}$

$\displaystyle dt = \frac{1}{\cos^2 \frac{\theta}{2}}\frac{d\theta}{2}$

$\displaystyle \int \frac{dt}{\sqrt{1 + t^2}} = \int \frac{\cos \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}\frac{d\theta}{2}$

$\displaystyle \int \frac{dt}{\sqrt{1 + t^2}} = \int \frac{1}{\cos \frac{\theta}{2}}\frac{d\theta}{2} = \int \frac{1}{\cos\theta}d\theta$

$\displaystyle \int \frac{dt}{\sqrt{1 + t^2}} = {\rm arcsinh} t$

$\displaystyle \int \frac{1}{\cos\theta}d\theta = {\rm arcsinh} \tan \theta$