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I'm supposed to prove the following by using the hyperbolic sine double angle identity: $\sinh(2x)=2\sinh x\cosh x$ and archsinhx formula: $\ln(x+\sqrt{x^2+1})$ but can't seem to figure out the steps.

Prove: $\int\sqrt{x^2+a^2}\,dx = \frac{a^2}2 \ln\left(x+\sqrt{x^2+a^2}\right) + \frac{x}2 \sqrt{x^2+a^2} + C$

So far what I have is:

$$\int\sqrt{x^2+a^2}dx = a^2\int\cosh^2udu\ \ (\text{substitute }x=\operatorname{asinh}u)$$ $$ = \int(1+\sinh^2u)\ du = \int(\cosh(2u)-\cosh^2u)\ du = a^2\left(\frac{u}2-\frac14\sinh(2u)\right)$$

After that I'm just stuck. I've tried substituting $x$ back but no matter how I go about it, I can't seem to drive the proof. Can someone give me a nudge?

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1 Answer 1

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Use the double angle identity for $\cosh$. We have $$\cosh 2u=2\cosh^2 u-1,$$ and therefore $$\cosh^2 u=\frac{1}{2}(1+\cosh 2u)$$ Integrate. We get $$\frac{u}{2}+\frac{1}{4}\sinh 2u+C=\frac{u}{2}+\frac{1}{2}\cosh u\sinh u +C.$$ Now the back substitution should go nicely.

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  • $\begingroup$ I did get something similar to what you had but my problem was exactly the back substitution part. What I'm doing is basically replacing all occurrences of u with $u = archsinh(\frac{x}{a}) = ln(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+a^2})$. Does that sound right? $\endgroup$
    – James
    Jun 26, 2013 at 4:11
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    $\begingroup$ Sort of. Except that you want to replace $\sinh u$ directly by $\frac{x}{a}$, and $\cosh u$ by almost directly by $\sqrt{1+(x/a)^2}=\frac{1}{a}\sqrt{a^2+x^2}$. It is only the naked $u$ in $\frac{u}{2}$ that you want to replace by the $\text{arcsinh}$. $\endgroup$ Jun 26, 2013 at 4:26
  • $\begingroup$ I understand why $sinhu$ is directly replaced by $\frac{x}{a}$ since $sinh(arcsinh(\frac{x}{a})) = \frac{x}{a}$ but I can't seem to figure out how you replaced $coshu$ with that. I tried to expand $cosh(arcsinh(\frac{x}{a}))$ but can't arrive to what you got. I feel really dumb asking about this :( $\endgroup$
    – James
    Jun 26, 2013 at 4:45
  • $\begingroup$ $1+\sinh^2 u=\cosh^2 u$, it is something you mentioned. So $\cosh^2 u=1+(x/a)^2$. Take the square root, no ambiguity since $\cosh$ is positive. $\endgroup$ Jun 26, 2013 at 4:59

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