# Is This Way Correct for Solving this Integral

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?

• I am unable to correct the powers in mathjax due to some reason. Please help me with the edits – user671231 Jul 23 at 18:45
• @VedantChourey ok I'll try thanks – user671231 Jul 23 at 18:50
• Your method is correct but it seems to be too round about. I can't see any added advantage in your approach. – Anurag A Jul 23 at 19:51
• @AnuragA I was just trying some things and this came to my mind. I thought it might helpful in other places where maybe the numerator may have been more complex and difficult to solve using trig identities. Is there a name for this method though? – user671231 Jul 23 at 19:53

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.