Trig substitution integral

  • Thread starter iRaid
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  • #1
iRaid
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Homework Statement


[tex]\int \frac{dx}{\sqrt{x^{2}+16}}[/tex]


Homework Equations





The Attempt at a Solution


[tex]x=4tan\theta[/tex] [tex]dx=4sec^{2}\theta d\theta[/tex]
Therefore:
[tex]\int \frac{4sec^{2}\theta d\theta}{\sqrt{16tan^{2}\theta +16}} = \int \frac{sec^{2}\theta d\theta}{\sqrt{tan^{2}\theta+1}}[/tex]
[tex]\int \frac{sec^{2}\theta d\theta}{sec\theta} = \int sec\theta d\theta[/tex]
[tex]=ln|sec\theta+tan\theta| = ln|sec\frac{\sqrt{x^{2}+16}}{4}+tan\frac{x}{4}|[/tex]

Wolfram is getting a hyperbolic sine function so idk what is wrong (we've never talked about hyperbolic functions in class, so I don't think that's the answer)
 

Answers and Replies

  • #2
LCKurtz
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You are OK at ##\ln|\sec\theta+\tan\theta|##. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

Don't worry about the hyperbolic functions. Integrals like this can be worked either way.
 
  • #3
iRaid
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You are OK at ##\ln|\sec\theta+\tan\theta|##. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

Don't worry about the hyperbolic functions. Integrals like this can be worked either way.

Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

Thanks for the help
 
  • #4
LCKurtz
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Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

Thanks for the help

They can be easier, but in this case I don't think it makes much difference. Try ##x = 4\sinh t## on it and see what you think. Remember the basic hyperbolic identity is ##\cosh^2t - \sinh^2 t = 1##.
 

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