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

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