Integral of 1/1+sinx

**animesh271094** · August 23rd 2011, 08:39 AM

How do you integrate this:
$\int \frac {1}{1+sinx}dx$

I tried expanding the '1' as $sin^2 x+cos^2 x$
And then i tried multiplying numerator and denominator by $cos x$ .. But it was of no use..

What do you do generally, when you have a trigonometric equation in the denominator??

**Plato** · August 23rd 2011, 08:47 AM

Originally Posted by animesh271094

How do you integrate this:
$\int \frac {1}{1+sinx}dx$

It is true that $\frac {1}{1+sinx}=\frac{1-\sin(x)}{\cos^2(x)}=\sec^2(x)+\frac{-\sin(x)}{\cos^2(x)}$

**animesh271094** · August 23rd 2011, 09:08 AM

Alright.. thanx..

**Siron** · August 23rd 2011, 09:45 AM

Another way is with the Weierstrass substitution:
Let $t=\tan \left(\frac{x}{2}\right)$ and so the integral becomes:
$\int \frac{\frac{2}{1+t^2}}{1+\frac{2t}{1+t^2}}dt$
$=\int \left(\frac{2}{1+t^2}\cdot \frac{1+t^2}{1+t^2+2t}\right)dt$
$=\int \frac{2dt}{t^2+2t+1}dt=\int \frac{2dt}{(t+1)\cdot (t+1)}$

Now split in partial fractions.

**Krizalid** · August 23rd 2011, 11:12 AM

Why partial fractions when having $t^2+2t+1=(t+1)^2$ ?

**Siron** · August 23rd 2011, 12:12 PM

Yes, offcourse! I was to busy with partial fractions I didn't noticed that, thanks

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