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

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Answer 1 · 2015-01-29T09:49:56

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#