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Basic indefinite integral

  • #1
Hi guys, I don't even know where to begin with this question.

Find the following indefinite integral:

[tex] \int x.arsinh (x^2) dx [/tex]

Thanks very much for any help, its much appreciated.
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
13
Well, first you'll need a u-substitution and then integration by parts. The u-sub is simple, so I'll skip that part. For the integration by parts, try putting u=arcsin(x) and dv=dx.
 
  • #3
Thanks for that, I got the following final answer:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?
 
Last edited:
  • #5
33,262
4,962
Thanks for that, I got the following final answer:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?
The substitution should be u = arcsinh(x). There is no arsinh function.
 
  • #6
Char. Limit
Gold Member
1,204
13
Woops! I thought that read arcsin(x)! My bad.
 
  • #7
Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?
 
  • #8
33,262
4,962
Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?
Did your teacher also write arcsin and arccos as arsin and arcos?
 
  • #9
No they were written as arcsin and arccos. It was just the hyperbolic ones that were written without a c.
 
  • #10
Char. Limit
Gold Member
1,204
13
arsinh and arcosh are the correct forms; they stand for "area hyperbolic sine" and "area hyperbolic cosine", respectively.
 
  • #11
33,262
4,962
arsinh and arcosh are the correct forms; they stand for "area hyperbolic sine" and "area hyperbolic cosine", respectively.
Well, ya learn something every day.
 

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