- #36

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and even what @PeroK presented in post 12 takes a little algebra to compute, starting with ## x=\cosh{u}=\frac{e^u+e^{-u}}{2} ##.

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- Thread starter askor
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- #36

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- #37

askor

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##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.

- #38

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Using the above definitions what are the derivatives of ##\sinh x## and ##\cosh x##?

##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.

What is ##\cosh^2 x## in terms of ##\sinh^2 x##?

In terms of integration, you use them the same way you use the trig functions, by substitution. E.g.:

$$x = \cosh u, \ \ dx = \frac{d}{du} (\cosh u) du$$

- #39

I think finding ##dx## helps so you need to find the derivative of ##\cosh x##. I learned it from a table of derivatives and integrals, which contained also ##\cosh x##.

##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.

- #40

octopus26

- 9

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Wolfram AlphaHow do you integrate ##\frac{1}{\sqrt{x^2 + 3x + 2}} dx##?

I had tried using ##u = x^2 + 3x + 2## and trigonometry substitution but failed.

Please give me some clues and hints.

Thank you

mentor note: moved from a non-homework to here hence no template.

- #41

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The hyperbolic trig approach is neater, but to proceed with the above form use partial fractions.##\int \frac{du}{1 - u^2}##

- #42

Grasshopper

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With you have there, why couldn’t a second substitution be made, using ##1 - sin^2w = cos^2w##?The hyperbolic trig approach is neater, but to proceed with the above form use partial fractions.

I’m sure I’m missing something.

- #43

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That would be going back to what we had earlier, integrating sec.With you have there, why couldn’t a second substitution be made, using ##1 - sin^2w = cos^2w##?

I’m sure I’m missing something.

Do you see how to solve it using partial fractions?

- #44

Grasshopper

Gold Member

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Wait, lol sorry I see now. It was late I was not thinking clearly. But yeah this is one of the easier partial fractions.That would be going back to what we had earlier, integrating sec.

Do you see how to solve it using partial fractions?

- #45

chwala

Gold Member

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what is the final solution? did you really get it?Wait, lol sorry I see now. It was late I was not thinking clearly. But yeah this is one of the easier partial fractions.

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