Confusing Trig/rational Integral

In summary, the conversation revolved around solving the integral \int (sin(t)-cos(t)) \sqrt{cos^2(t)-sin^2(t)} dt using trigonometric identities and substitutions. It was determined that the correct identity to use was cos^2(t) - sin^2(t) = cos(2t). One method involved integrating by parts, while another involved using substitutions. The final solution was found to be -\frac{\sqrt{2}}{8}\sinh \left(2 \ \mbox{arg}\cosh\left(\sqrt{2}\cos t\right)\right)+\frac{\sqrt{2}}{4}\mbox{arg}\cosh\left(\
  • #1
Chaz706
13
0
[tex] \int (sin(t)-cos(t)) \sqrt{cos^2(t)-sin^2(t)} dt [/tex]

Is there a trig idendity I can use? I've distributed that root to both terms to get:
[tex] \int sin(t) \sqrt{cos^2(t)-sin^2(t)} dt -[/tex] [tex] \int cos(t) \sqrt{cos^2(t)-sin^2(t)} [/tex]

If I take one of the terms and integrate by parts, I'm trying to put [tex] u=\sqrt{cos^2(t)-sin^2(t)} [/tex] and [tex]dv= sin(t) [/tex] or [tex]dv= cos(t) [/tex] but that ugly root's derivative appears inside the [tex] \int vdu [/tex] part.

Is there a trig identity I'm missing, or some other tactic I could use? or could this just be really really ugly math?
 
Physics news on Phys.org
  • #2
Its either

[tex] cos^2(t) - sin^2(t) = cos(2t) [/tex]**

or

[tex] sin^2(t) - cos^2(t) = cos(2t) [/tex]
 
  • #3
Whozum:
[tex] cos^2(t) - sin^2(t) = cos(2t) [/tex] is the correct identity (and thanks! It may just help!)

Your other identity: [tex] sin^2(t) - cos^2(t) = cos(2t) [/tex] is erroneous, but not by much. [tex] sin^2(t) - cos^2(t) [/tex] yields [tex] -cos(2t) [/tex]

This could actually help as well, however. Thank you for posting both (even if one was inaccurate).
 
  • #4
You needn't that identity.

[tex] \int \left(\sin t-\cos t\right)\sqrt{\cos^{2}t-\sin^{2}t} \ dt = \int \sin t\sqrt{2\cos^{2}t-1} \ dt -\int \cos t\sqrt{1-2\sin^{2}t} \ dt [/tex]

Can u take it from here (HINT:eek:bvious substitutions necessary) ?

Daniel.
 
  • #5
That would have been much easier dex, I think...
 
  • #6
Well it was one of the two, I justw anst sure which one. dex's way is easier though its basically the same thing.
 
  • #7
The Method is substitution, but then what's u? I'm guessing the basic trig function outside the root is du.

But when I derive the function within that root, I get something different. The derivative of either one ends up to be [tex]-4cos(t)sin(t) [/tex].

Am I choosing the wrong U? I'm integrating with respect to t, and simply inserting the extra parts onto by basic trig function isn't exactly kosher.
 
  • #8
Okay,let's take the first,i'll leave with the second,which is basically similar (maybe other substitution).

[tex] I=\int \sin t\sqrt{2\cos^{2}t-1} \ dt [/tex] (1)

becomes after the substitution [itex] \cos t=u;\sin t \ dt= -du [/itex] (2)

[tex] I=-\int \sqrt{2u^{2}-1} \ du [/tex](3)

Make the substitution

[tex]\sqrt{2} \ u=\cosh v;du=\frac{1}{\sqrt{2}} \sinh v \ dv [/tex](4)

Then

[tex] I=-\frac{\sqrt{2}}{2}\int \sinh^{2}v \ dv=-\frac{\sqrt{2}}{4}\int (\cosh 2v-1) \ dv=-\frac{\sqrt{2}}{8}\sinh 2v+\frac{\sqrt{2}}{4}v +C [/tex] (5)

And now invert the 2 substitutions

[tex] I=-\frac{\sqrt{2}}{8}\sinh \left(2 \ \mbox{arg}\cosh\left(\sqrt{2}\cos t\right)\right)+\frac{\sqrt{2}}{4}\mbox{arg}\cosh\left(\sqrt{2}\cos t\right) + C [/tex] (6)

I'm sure the other one will be simpler.

Daniel.
 
Last edited:
  • #9
Didn't know that if you took [tex] cos(t) = u [/tex] that you could allow [tex] u^2 [/tex] to simply be [tex] cos^2(t) [/tex].

Of course, I didn't think about that either.
 
  • #10
The other way around is incorrect,of course,but i simply squared.

Daniel.
 

Related to Confusing Trig/rational Integral

What is a confusing trigonometric and rational integral?

A confusing trigonometric and rational integral is an integral that involves both trigonometric and rational functions, making it difficult to solve using traditional integration techniques. These types of integrals often require special techniques or substitutions to solve.

Why are trigonometric and rational integrals considered difficult?

Trigonometric and rational integrals can be difficult because they involve multiple functions and require a combination of algebraic and trigonometric manipulation to solve. Additionally, they may not have a straightforward antiderivative, making it challenging to find a solution.

What are some common trigonometric and rational integrals?

Some common trigonometric and rational integrals include integrals of trigonometric functions raised to a power, rational functions multiplied by trigonometric functions, and inverse trigonometric functions. Examples include ∫sin^3(x)dx, ∫x/(x^2 + 1)dx, and ∫arctan(x)/x dx.

How can I solve a confusing trigonometric and rational integral?

To solve a confusing trigonometric and rational integral, you can use techniques such as trigonometric substitutions, partial fractions, or integration by parts. It is also helpful to have a good understanding of trigonometric identities and algebraic manipulation.

Are there any tips for solving confusing trigonometric and rational integrals?

Yes, here are a few tips for solving confusing trigonometric and rational integrals:
- Look for patterns and use common trigonometric identities to simplify the integral.
- If possible, try to rewrite the integral in terms of a single trigonometric function.
- When using trigonometric substitutions, make sure to choose the appropriate substitution that will simplify the integral.
- Be careful with signs and don't forget to include the constant of integration in your final answer.
- Practice, practice, practice! The more you work on these types of integrals, the more familiar you will become with the techniques and shortcuts.

Similar threads

Replies
12
Views
976
  • Introductory Physics Homework Help
Replies
8
Views
636
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
241
  • Calculus and Beyond Homework Help
Replies
15
Views
812
  • Introductory Physics Homework Help
Replies
17
Views
498
  • Introductory Physics Homework Help
Replies
16
Views
1K
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
628
  • Calculus and Beyond Homework Help
Replies
4
Views
286
Back
Top