How to integrate $$\int \frac{1}{g-kv^2}~dv$$ where $v$ is velocity and $g,k$ are known constants.
I have been stuck for a while on this and have now turned to SE for help any suggestions?
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1Could you do the integral if $g$ and $k$ were both $1$? I.e., do you know how to integrate $\int{1\over1-v^2},dv$? – Barry Cipra Mar 24 '15 at 15:47
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Partial fractions then logs? – Mechin Mar 24 '15 at 15:50
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As a starter: Factor the denominator, apply the partial fractions decomposition, and then integrate each term. You can do the rest. – Meshal Mar 24 '15 at 15:50
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What do you mean factor the denominator? – Mechin Mar 24 '15 at 15:54
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Like take out $k$ and then make it the product of two linear terms in $v$? – Mechin Mar 24 '15 at 15:54
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@Mechin Exactly. – Meshal Mar 24 '15 at 15:55
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@Mechin, yes, partial fractions then logs. To do partial fractions, you need to factor the denominator (which is what kobe does in an answer). – Barry Cipra Mar 24 '15 at 15:57
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@Mechin I provided an alternative methodology that might be useful to you both here and down the road. It uses hyperbolic trigonometric substitution, which is analogous to trigonometric substitution. Let me know how I can make my answer better. I just want to give you the best possible answer I can give you. – Mark Viola Mar 24 '15 at 18:01
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I'm assuming $g, k > 0$. Let $\omega^2 = g/k$. Then
\begin{align}\int \frac{1}{g - kv^2}\, dv &= \frac{1}{k}\int \frac{1}{\omega^2 - v^2}\, dv\\ & = \frac{1}{2\omega k}\int \left(\frac{1}{\omega - v} + \frac{1}{\omega + v}\right)\, dv\\ & = \frac{1}{2\omega k}(-\log|\omega - v| + \log|\omega + v|) + C\\ & = \frac{1}{2\omega k}\log \left|\frac{\omega + v}{\omega - v}\right| + C\\ & = \frac{1}{2\sqrt{gk}}\log\left|\frac{\sqrt{g/k} + v}{\sqrt{g/k} - v}\right| + C. \end{align}
kobe
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This solution assumes $g/k>0$. It works but involves a choice of branch cut if $g/k<0$. – Ian Mar 24 '15 at 15:53
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@abel, I think $gk\gt0$ is a better way to say it, because $g/k$ is not defined (as a real number) when $k=0$. – Barry Cipra Mar 24 '15 at 16:09
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Kobe provided an excellent solution using partial fraction expansions. So, here, I'll use a different approach - hyperbolic trigonometric substitution.
To that end, let $I$ be the indefinite integral of interest, namely $I=\int \frac{dv}{g-kv^2}$. Now, let's make the substitution $v=\sqrt{\frac{g}{k}}\tanh x$. Then, we have $dv=\sqrt{\frac{g}{k}} \text{sech}^2 x dx$ and
$$\begin{align} I & = \int \frac{dv}{g-kv^2} \\ &= \int \frac{\sqrt{\frac{g}{k}} \text{sech}^2 x \; dx}{g\left(1-\tanh^2x\right)} \\ & =\frac{1}{\sqrt{gk}} x +C\\ & =\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{\frac{k}{g}}v\right) +C \end{align}$$
which is an acceptable final result. If one prefers, one may use identity $\tanh^{-1} x = \log (\sqrt{\frac{1+x}{1-x}})$ to yield an alternative expression for $I$ as
$$\begin{align} I & =\frac{1}{\sqrt{gk}}\log\left(\sqrt{\frac{\sqrt{\frac{g}{k}}+v}{\sqrt{\frac{g}{k}}-v}}\right) +C \\ & =\frac{1}{2\sqrt{gk}} \log \left( \frac{ \sqrt{\frac{g}{k}}+v }{ \sqrt{\frac{g}{k}}-v}\right) +C \end{align}$$
which recovers the aforementioned result!
Mark Viola
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