Hyperbolic Substituion. I am wrong?

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In summary, the conversation discusses the integration of sinh(2x) cosh(2x) dx and the use of substitutions to simplify the integral. It is shown that both methods of substitution, u = sinh(2x) and u = cosh(2x), lead to the same result with a difference in constant. The conversation also mentions the use of WolframAlpha to check the result of the integral.
  • #1
flyingpig
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Homework Statement

[tex]\int \;\sinh(2x) \cosh(2x) dx[/tex]

The Attempt at a Solution



I let u = sinh(2x), du = 2cosh(2x)dx

Integrating I shuold get[tex]\frac{1}{4} sinh^2 (2x) + C[/tex]

But http://www.wolframalpha.com/input/?i=Integrate[cosh%282x%29%28sinh%282x%29%29%2Cx]

says I need to let u = 2x first. Why? I don't think i am wrong, my algebra looked correct.
 
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  • #2
No, you don't need to do that, you just can.

If you let u= 2x, then, of course, dx= (1/2) du so your integral becomes
[tex]\frac{1}{2}\int sinh(u)cosh(u)du[/tex]
Then let v= sinh(u) so that dv= cosh(u)du and the integral becomes
[tex]\frac{1}{2}\int v dv= \frac{1}{4}v^2+ C= \frac{1}{4}sinh^2(u)+ C= \frac{1}{4}sinh^2(2x)+ C[/tex]
just as you have.
That's just using two consecutive substitutions rather than combining them as you did.

Of course, you could do it the other way two: let u= cosh(2x) so that du= 2sinh(x)dx and the integral becomes
[tex]\frac{1}{2}u^2du= \frac{1}{4}cosh^2(2x)+ D[/tex]

Because [itex]cosh^2(2x)= sinh^2(2x)+ 1[/itex] that is exactly the same thing with a difference constant.

I believe your wolframalpha reference gives the integral as
[tex]\frac{1}{8}cosh(4x)+ Constant[/tex]
which looks very different but is actually the same as your result.

Remember that
[tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/tex]

Squaring that:
[tex]sinh^2(x)= \frac{(e^x)^2- 2e^xe^{-x}+ (e^{-x})^2}{4}= \frac{e^{2x}- 2+ e^{-2x}}{4}[/tex]
[tex]= \frac{1}{2}\frac{e^{2x}+ e^{-2x}}{2}- \frac{1}{2}= \frac{1}{2}cosh(2x)- \frac{1}{2}[/tex]
so that
[tex]\frac{1}{4}sinh^2(2x)= \frac{1}{8}cosh(4x)- \frac{1}{8}[/tex]
differing from the original form only by a constant.
 
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  • #3
Oh right, we are doing an indefinite integral here.
 

1. What is hyperbolic substitution?

Hyperbolic substitution is a mathematical technique used to simplify integrals involving trigonometric functions. It involves replacing trigonometric functions with hyperbolic functions, which have simpler derivatives.

2. How is hyperbolic substitution used?

Hyperbolic substitution is used to solve integrals involving trigonometric functions, particularly when the integrand contains a square root of a quadratic expression.

3. Can hyperbolic substitution be used for any integral?

No, hyperbolic substitution is only useful for certain types of integrals involving trigonometric functions. It is not a universal method for solving all integrals.

4. What are the benefits of using hyperbolic substitution?

Hyperbolic substitution can simplify integrals and make them easier to solve. It can also help to avoid using more complicated integration techniques such as integration by parts or partial fractions.

5. Are there any limitations or drawbacks to using hyperbolic substitution?

Hyperbolic substitution can only be used for specific types of integrals, so it is not always applicable. It also requires a good understanding of hyperbolic functions and their derivatives, which may be challenging for some individuals.

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