# Reciprocal of Trig - Integral

1. Feb 21, 2006

### Aresius

Hi I have a proof i'm doing

$$\int \frac{1}{1+\sin(x)}dx$$

I know that the answer i'm looking for is

$$\frac{\sin(x) - 1}{\cos(x)}$$
and then
$$\tan(x) - \sec(x)$$

I have tried integration by parts making
$$u = (1+\sin(x))^{-1}$$ and $$dv = dx$$

Eventually I get an answer that contains an ln and an unsolvable integral. I have been at this for 2 hours, can anyone give me a hint or a push in the right direction?

Last edited: Feb 21, 2006
2. Feb 21, 2006

### Aresius

And I will be fixing my latex right now...

3. Feb 21, 2006

### TD

It's possible to do a t-substitution (t = tan(x/2)) but you can also use a little trick.

$$\int {\frac{1} {{1 + \sin x}}dx} = \int {\frac{{1 - \sin x}} {{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} dx = \int {\frac{{1 - \sin x}} {{1 - \sin ^2 x}}} dx = \int {\frac{{1 - \sin x}} {{\cos ^2 x}}} dx$$

Now split the integral (so the fraction...) in two: the first part should be a standard integral and the second will go easily using an obvious substitution.

4. Feb 21, 2006

### Aresius

I tried that t thing and it seemed too complicated for the level we are currently at. The trick worked, and I didn't have to use any substitution - just rearrange the sinx/(cosx)^2 to secxtanx which is an easy integral.

Thanks for the fast response.

5. Feb 21, 2006

### TD

You're welcome.

If you'd like to try it using the t-substitution too, post your work or ask for help