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

In summary, the user is struggling with the problem of integrating 1/(1+sin(x)), with the goal of reaching an answer of (sin(x)-1)/cos(x) or tan(x)-sec(x). They have attempted integration by parts and using a t-substitution, but have not been successful. Another user suggests using a trick to rearrange the integral into a more manageable form, and the original poster is able to solve the problem using this method. They thank the responder for their quick help.
  • #1
Aresius
49
0
Hi I have a proof I'm doing

[tex]
\int \frac{1}{1+\sin(x)}dx
[/tex]

I know that the answer I'm looking for is

[tex]
\frac{\sin(x) - 1}{\cos(x)}
[/tex]
and then
[tex]
\tan(x) - \sec(x)
[/tex]

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

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:
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  • #2
And I will be fixing my latex right now...
 
  • #3
It's possible to do a t-substitution (t = tan(x/2)) but you can also use a little trick.

[tex]
\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
[/tex]

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
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
Aresius said:
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 :smile:
 

What is the reciprocal of trigonometric functions?

The reciprocal of a trigonometric function is the inverse of the function. It is obtained by interchanging the x and y coordinates of a point on the graph of the original function.

What is an integral in trigonometry?

In trigonometry, an integral is a mathematical operation that determines the area under a curve on a graph. It is represented by the symbol ∫ and is used to find the exact value of a function over a given interval.

How do you find the reciprocal of a trigonometric function?

The reciprocal of a trigonometric function can be found by interchanging the x and y coordinates of a point on the graph of the original function. For example, the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

What is the relationship between integrals and trigonometric functions?

Integrals and trigonometric functions are closely related, as the integral of a trigonometric function is used to find the area under the curve on a graph. Trigonometric functions are also used to model periodic phenomena, which can be integrated to find the total value over a given time period.

Why are integrals important in trigonometry?

Integrals are important in trigonometry because they allow us to calculate the exact value of a function over a given interval. This is useful in various applications, such as calculating the area of a shape or finding the displacement of an object over time. Integrals also play a key role in calculus, which is used in many scientific and engineering fields.

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