Question

How to find the integral of `1/cosx`?

Answer 1

You need to use the bioche rule

Règle de Bioche (In French)

I don't find it in English, but trust me it's VERY usefull.

The rules tell, you need to do `t = tan(x/2)`

`cos(x) = (1-t^2)/(1+t^2) => 1/cos(x) = (1+t^2)/(1-t^2)`

`dx = 2/(1+t^2) dt`

So now we have :

`int(1+t^2)/(1-t^2)*2/(1+t^2) dt = 2int1/(1-t^2)`

`1/(1-t^2)` is exactly the derivate of `arctanh(t)`

How to remember this derivate easily ?

`theta = arctanh(t)`

`=>tanh(theta) = t`

Derivate both side

`=>d theta(1-tanh^2(theta)) = 1`

`=>d theta =1/(1-tanh^2(theta))`

but `tanh (theta)= t`

Finally :

`=>d theta = 1/(1-t^2)`

You can do this with arccos, arcsin, arctan... using pythagore

So the integral is :

`=> [arctanh(t)]+C`

Substitute back for `t = tan(1/2x)`

`=>[arctanh(tan(1/2x))]+C`