What is the trick to solving the integral of 1/(1+sin(x))?

[Aresius]

Hi I have a proof I'm doing

<br /> \int \frac{1}{1+\sin(x)}dx<br />

I know that the answer I'm looking for is

<br /> \frac{\sin(x) - 1}{\cos(x)}<br />
and then
<br /> \tan(x) - \sec(x)<br />

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

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?

 

[Aresius]

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

 

[TD]

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

<br /> \int {\frac{1}<br /> {{1 + \sin x}}dx} = \int {\frac{{1 - \sin x}}<br /> {{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} dx = \int {\frac{{1 - \sin x}}<br /> {{1 - \sin ^2 x}}} dx = \int {\frac{{1 - \sin x}}<br /> {{\cos ^2 x}}} dx<br />

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.

 

[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.

 

[TD]

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.

You're welcome.

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