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?

  • $\begingroup$ I am unable to correct the powers in mathjax due to some reason. Please help me with the edits $\endgroup$
    – user671231
    Commented Jul 23, 2019 at 18:45
  • $\begingroup$ @VedantChourey ok I'll try thanks $\endgroup$
    – user671231
    Commented Jul 23, 2019 at 18:50
  • 1
    $\begingroup$ Your method is correct but it seems to be too round about. I can't see any added advantage in your approach. $\endgroup$
    – Anurag A
    Commented Jul 23, 2019 at 19:51
  • $\begingroup$ @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? $\endgroup$
    – user671231
    Commented Jul 23, 2019 at 19:53

1 Answer 1


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.

  • 2
    $\begingroup$ Thanks for the answer but I know how to solve the Integral I want to know if my method is right or not and that is there a name for this method. What you've done is also basically dealing with Trigonometric formulas $\endgroup$
    – user671231
    Commented Jul 23, 2019 at 19:02

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