Question

Why can't you integrate `sqrt(1+(cosx/-sinx)^2`?

Answer 1

The answer is: `ln|tan(x/2)|+c`

First of all it's useful to "change" a little the function:

`y=sqrt(1+(cosx/-sinx)^2)=sqrt(1+(cos^2x)/(sin^2x))=sqrt((sin^2x+cos^2x)/(sin^2x))=sqrt(1/(sin^2x))=1/sinx`.

I used the first fondamental relation of trigonometry: `sin^2x+cos^2x=1`

So we have to integrate: `int1/sinxdx`.

Before integrate this function, it is useful to remember the parametric formula of sinus, that says:

`sinx=(2t)/(1+t^2)`, where `t=tan(x/2)`.

Now the integral will be done with the method of substitution:

`tan(x/2)=trArrx/2=arctantrArrx=2arctantrArrdx=2(1/(1+t^2))dt`.

Our integral becoms:

`int1/sinxdx=int1/((2t)/(1+t^2))2(1/(1+t^2))dt=int(1+t^2)/(2t)2(1/(1+t^2))dt=int1/tdt=ln|t|+c=ln|tan(x/2)|+c`