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Evaluate $$\int\frac{dy}{\sqrt{169 + y^2}}$$

I have solved the problem, but don't seem to be getting the right answer. I get $\ln|\sqrt{13+y}+y|+C$ as the answer, and it is not the answer, how can I solve this?

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  • $\begingroup$ What is your answer, and what is the given answer? If you've done everything right, the two answers might just differ by an additive constant. $\endgroup$ – Tunococ Aug 2 '14 at 2:56
  • $\begingroup$ I don not know the answer given, but my answer is ln|13+y‾‾‾‾‾‾‾√+y|+C $\endgroup$ – brian Aug 2 '14 at 2:59
  • $\begingroup$ If you know hyperbolic sine, there is any easy substitution. Otherwise double check how you missed out $\sqrt{169+y^2}$ and wrote $\sqrt{13+y}$ in your workings. $\endgroup$ – Macavity Aug 2 '14 at 3:07
  • $\begingroup$ okay, I think the answer actually might be ln|(√(169+y^2)/13)-(y/13)+C $\endgroup$ – brian Aug 2 '14 at 3:10
  • $\begingroup$ or + (y/13) i think $\endgroup$ – brian Aug 2 '14 at 3:10
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There is another way to do this, one that I find highly interesting. Credit must go to @juantheron, whose post in another question first mentioned this.

Start by defining $\displaystyle x = \sqrt{y^2+a^2} \implies x^2 = y^2 + a^2 \implies x\,dx = y\,dy \implies \frac{dx}{y} = \frac{dy}{x}$

Using the simple algebra of proportions, we can see that $\displaystyle\frac{dx}{y} = \frac{dy}{x} = \frac{dx+dy}{x+y} = \frac{d(x+y)}{x+y}$.

Hence the original integral is $\displaystyle \int \frac{dy}{x} = \int \frac{d(x+y)}{x+y} = \ln |x+y| + C = \ln|y + \sqrt{y^2 + a^2}| + C$.

In your case, $\displaystyle a=13$.

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Use the substitution $x=13\tan{u}$, then $dx=13\sec^2{u} \ du$ \begin{align} \int\frac{dx}{\sqrt{169+x^2}} &=\int\frac{13\sec^2{u} \ du}{\sqrt{169+169\tan^2{u}}}\\ &=\int\sec{u} \ du\\ &=\ln|\sec{u}+\tan{u}|+c\\ &=\ln\left|\frac{x}{13}+\sqrt{1+\left(\frac{x}{13}\right)^2}\right|+c \end{align} Alternatively, we can use the hyperbolic substitution $x=13\sinh{u}$, then $dx=13\cosh{u} \ du$. Thus \begin{align} \int\frac{dx}{\sqrt{169+x^2}} &=\int\frac{13\cosh{u} \ du}{\sqrt{169+169\sinh^2{u}}}\\ &=\int\frac{13\cosh{u} \ du}{13\cosh{u}}\\ &=\operatorname{arcsinh}\left(\frac{x}{13}\right)+c \end{align} These 2 answers are equal.

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    $\begingroup$ @brian: If the official answer is $\ln|y+\sqrt{169+y^2}|+C$, that is "the same" as the one above, because the constant can be absorbed into the constant of integration. $\endgroup$ – André Nicolas Aug 2 '14 at 3:08
  • $\begingroup$ I honestly have no clue $\endgroup$ – brian Aug 2 '14 at 3:13
  • $\begingroup$ @Macavity Thanks for pointing out that blunder. I have edited it. $\endgroup$ – SuperAbound Aug 2 '14 at 3:19

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