Integral of inverse trig or inverse hyperbolic

  • Thread starter USN2ENG
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  • #1
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Homework Statement



∫5/(4x√(9-16x2)dx

Homework Equations



I am pretty sure this is in the form of ∫du/(u√(a2-u2)

The Attempt at a Solution



setting u=4x a=3 and du=4dx so 1/4du=dx I get:

-5/12 sech-1(4x/3) + C

Is this right or am I using the wrong definition? Just trying to check my answers

Thanks for any help
 

Answers and Replies

  • #2
HallsofIvy
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Yes, that is correct.
 
  • #3
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Thanks, I really appreciate it!

One quick follow-up question to anyone who can help:

Some of the integral definitions involving hyperbolic inverse functions call for if a>u or u>a. I know that dealing with a definite integral we just use the limits of integration to figure that out, but what if we are dealing with an indefinite integral? How do you know then?
 

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