I was recently trying this question
$$ \int \frac {\sin x dx}{(1+\sin x)} =$$
I know how to solve it using Half Angle Tangent Substitutions however I tried another method and want to know whether it's correct or not-
I took $$ \ln(1+\sin x)= t$$
I got $$ \frac {\cos x dx}{(1+\sin x)} =dt$$
Then I converted the Integral from the $~x~$ world to the $~t~$ world and converted $~\cos x~$ in terms of $~t~$ using my original substitution.
Now I broke the original integral into $~2~$ parts by adding and subtracting $~1~$ in the numerator.
The first Integral was a simple Integral of a constant term and I used the substitution in the second integral. I got the following
$$ \int \frac {e^{-t} dt}{(2e^{-t}-1)^{1/2}} =$$ Then I used the substitution $$ 2e^{-t}-1=k^{2}$$ and got the answer
Is this method good to go or am I doing something wrong. Also I found this method to particularly helpful in this case as I don't like to deal with half angle Substitutions. Is there a name for this method?
The integrand is the same as $$\frac{(1+\sin x)-1}{1+\sin x} = 1-\frac{1}{1+\sin x} = 1-\frac{1-\sin x}{\cos^2 x} = 1-\sec^2 x +\sec x \tan x$$ and each of this functions its easy to integrate.