Integration by Parts: Solving y = arcsinh(x)

In summary, to find y, which is equal to the integral of arcsinh(x) dx, we use integration by parts with the formula int(v.du)=uv-int(u.dv). We first perform a substitution, x = sinh(a), dx = cosh(a) da, and then integrate by parts.
  • #1
hex.halo
13
0

Homework Statement



Use integration by parts to find:

y=... if dy=arcsinh(x) dx

Homework Equations



int(v.du)=uv-int(u.dv)

The Attempt at a Solution



I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?
 
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  • #2
[tex]\int sinh^{-1}x dx[/tex]


well [itex]dv\neq sinh^{-1}x dx[/itex] since to find v you'd need to integrate that and well obviously you can't do that. So [itex]u=sinh^{-1}[/itex] and [itex]dv=1dx[/itex].
 
  • #3
A good general tip is to take very hyperbolic function and every trig function and learn how to differentiate it or integrate it. That way you don't have to remember the tables of functions. If you like memorising tables though do it that way, but do both. :smile: Just a really good but obvious hint I picked up recently that I thought might be useful.

It's generally a good idea to try and split arc functions into there [itex]\frac{1}{\text{trig function}}[/tex] equivalents first, I don't know if it's just me but that seems to work out better more often than not? Clearly here the substitution becomes much easier when you do this, but I find it's a good general rule..?
 
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  • #4
hex.halo said:
I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?

Hi hex.halo ! :smile:

You have to do a substitution first, and then integrate by parts!

Put x = sinha, dx = cosha da …

and you get … ? :smile:
 

1. What is Integration by Parts?

Integration by Parts is a method used in calculus to find the integral of a product of two functions. It is based on the product rule for differentiation and is useful when the integral cannot be easily solved by other methods.

2. How is Integration by Parts applied to solve y = arcsinh(x)?

To solve y = arcsinh(x) using Integration by Parts, we first express the function as a product of two functions: u = arcsinh(x) and dv = 1. Then, we use the integration by parts formula: ∫u dv = uv - ∫v du. After solving for the integral on the right side, we can find the value of the original integral.

3. What is the purpose of using Integration by Parts to solve y = arcsinh(x)?

The purpose of using Integration by Parts to solve y = arcsinh(x) is that it allows us to find the integral of the function without having to use other, more complicated methods. It is especially useful when the function involves a product of two functions that cannot be easily integrated.

4. Are there any limitations to using Integration by Parts to solve y = arcsinh(x)?

Yes, there are limitations to using Integration by Parts to solve y = arcsinh(x). This method is only applicable when the integral can be expressed as a product of two functions. It also requires a certain level of algebraic manipulation and may not always yield an exact answer.

5. Can Integration by Parts be used to solve other types of equations?

Yes, Integration by Parts can be used to solve a variety of equations. It is useful for finding the integral of a product of two functions, but it can also be used to solve integrals involving trigonometric functions, logarithmic functions, and inverse trigonometric functions.

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