Integrate 1/ ( sin x + sec x)

  • #1

Homework Statement


∫ 1 /( sin x + sec x) dx


Homework Equations





The Attempt at a Solution


∫ cos x / ( sin x + cos x ) style question
Tried ∫ cos x / (sin x . cos x + 1) dx
and uses sin 2x
tried substitutions
nothing seem to work

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
658
2
I'm really curious to see if anyone can get this one. I plotted the graph and it looks very strange.. Please post the solution when you find it, it would help me sleep at night.
 
  • #5
2,967
5
You can always reduce it to an integral with rational algebraic expressions with the substitution:
[tex]
t \equiv \tan \left( \frac{x}{2} \right)
[/tex]

[tex]
\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}
[/tex]
 
  • #6
349
1
You can always reduce it to an integral with rational algebraic expressions with the substitution:
[tex]
t \equiv \tan \left( \frac{x}{2} \right)
[/tex]

[tex]
\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}
[/tex]
man, i wish i knew how to do this back in the day. . .
 
  • #7
658
2
You can always reduce it to an integral with rational algebraic expressions with the substitution:
[tex]
t \equiv \tan \left( \frac{x}{2} \right)
[/tex]

[tex]
\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}
[/tex]
I can't see that being much easier because after playing with the algebra it's going to leave many t's with various higher degrees (up to 6th degree if my algebra is correct) and also in a fraction form. After plugging back in tan it's going to have some subs and then reduction formulas for those higher degrees.
 
  • #8
2,967
5
and this is exactly why the output from Wolfram Alpha looks that long. The only difference is that the imaginary unit is missing and instead of inverse hyperbolic, you'll have inverse trigonometric functions.

No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
You could try something like this:
\begin{align*}
\frac{1}{\sin x + \sec x} &= \frac{1}{\sin x + \sec x} \cdot \frac{\sin x - \sec x}{\sin x - \sec x} \\
&= \frac{\sin x - \sec x}{\sin^2 x - \sec^2 x} \\
&= \frac{\cos^2 x\sin x - \cos x}{\cos^2 x\ \sin^2 x - 1} \\
&= \frac{\cos^2 x}{\cos^2 x\ (1-\cos^2 x) - 1}\sin x - \frac{1}{(1-\sin^2 x)\sin^2 x - 1}\cos x
\end{align*}Then use a trig substitution for each of the two terms.
 
  • #10
658
2
No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.
Yeah I know what you mean, it's just that the integral just looked deceiving simple at first glance. :smile:
 
  • #11
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
Lol.. Oh my..

There has to be an easier way to get this one.
There is.

Use the fact that cos(-pi/4) = sin(-pi/4)=1/√2 .

[tex]\begin{aligned}
\sin x + \cos x &= \sqrt2 \left(\cos x\frac 1 {\sqrt2} + \sin x \frac 1 {\sqrt2}\right) \\
&= \sqrt2 (\cos x \cos(\pi/4) + \sin x \sin (\pi/4)) \\
&= \sqrt2 \cos(x-\pi/4)
\end{aligned}[/tex]

And thus [itex]1/(\sin x + \cos x) = \sec(x-\pi/4)/\sqrt 2[/itex]


So now all you have to do is find the integral of sec(x-pi/4) and then divide by √2.
 
  • #12
2,967
5
the only problem is that the denominator is [itex]\sin x + \sec x[/itex], and not [itex]\sin x + \cos x[/itex] as you were trying to simplify.
 

Related Threads on Integrate 1/ ( sin x + sec x)

  • Last Post
Replies
4
Views
1K
Replies
6
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
4
Views
31K
  • Last Post
Replies
4
Views
10K
  • Last Post
Replies
6
Views
14K
Top