Evaluate the following integral using the substitution $x=9\sinh^2\theta$ $$\int_0^1\sqrt{\frac{x+9}{x}}dx$$

So I have attempted this integral but, I've made a mistake somewhere and I cannot see anything wrong with what I've done, my workings are:

If $x=9\sinh^2\theta$, then $x+9 = 9\cosh^2\theta$ and $dx = 18\sinh\theta\cosh\theta \,d\theta$ $$\int \sqrt{\frac{9\cosh^2\theta}{9\sinh^2\theta}} \times 9\sinh\theta\cosh\theta \,d\theta$$ $$9\int \cosh^2\theta\, d\theta $$ Now using the identity that $\cosh^2\theta = \frac12 (\cosh2\theta +1)$, the integral becomes $$\frac92\int (\cosh2\theta +1 )\,d\theta$$ $$\frac92 \left[ \frac12 \sinh2\theta + \theta\right] = \frac92 \left[ \sinh\theta\cosh\theta + \theta\right]$$ The after messing around with some algebra, I got to $$\frac92 \left[ \frac19\sqrt{x} \sqrt{9+x} +\sinh^{-1}\sqrt{\frac{x}{9}}\right]^1_0$$ $$= \frac12 \left[\sqrt{10} + 9\sinh^{-1}\frac13 \right]$$

I have managed to get half of what the integral should be, my thoughts is that I've made a mistake somewhere in using an identity where Ii should have cancelled the half out. Any help would be great.

  • 3
    $\begingroup$ You substituted $dx$ incorrectly. $\endgroup$ – Peter Foreman Apr 17 at 20:39
  • 1
    $\begingroup$ That's annoyed me now that it was that simple. $\endgroup$ – H.Linkhorn Apr 17 at 20:40
  • $\begingroup$ Too many $9$s I guess. $\endgroup$ – Peter Foreman Apr 17 at 20:41
  • $\begingroup$ A factor $18$ became $9$, as you probably have already discovered. $\endgroup$ – egreg Apr 17 at 20:59

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