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I did a substitution on a DE and ended up with this:

$\int \frac{1}{4-v^2}dv$

I tried a trig substitute but things got a bit hairy. WolframAlpha recommended a far less intuitive substitution. I can't remember what it was.

What is the simplest way to solve this integral?

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    $\begingroup$ Partial fractions? $\endgroup$ – anon Jan 14 '14 at 1:42
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Hint: $\frac{1}{4-u^2}=\frac{\frac{1}{4}}{{2-u}}+\frac{\frac{1}{4}}{{2+u}}$

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  • $\begingroup$ Thanks. I hadn't even considered partial fractions. $\endgroup$ – Korgan Rivera Jan 14 '14 at 2:06
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Use partial fractions. Put $${1\over 4 -x^2} = {A\over 2-x} + {B\over 2+ x}.$$

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I'm surprised that with all these answers no one has mentioned hyperbolic trigonometric functions. If you know that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$ it may not surprise you to learn that $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$, and so:

$$\int \frac{1}{4-v^2}dv=\frac{1}{4}\int \frac{1}{1-(\frac{v}{2})^2} dv$$ substitude $u=\frac{v}{2}$ so $2 du=dv$

$$=\frac{1}{2} \int \frac{1}{1-u^2} du=\frac{1}{2} \tanh^{-1}(u)=\frac{1}{2} \tanh^{-1}(\frac{v}{2})$$

To prove the identities, try to mimic the proof that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$. That is, put $x=\tan(y)$ to find that $\frac{dy}{dx}=\cos^2(y)=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}$ so that integration with respect to x gives $y=\int \frac{1}{1+x^2}$. You'll need to prove the hyperbolic equivalent of $\sec^2(x)=1+\tan^2(x)$, $\text{sech}^2(x)=1-\tanh^2(x)$, which is trivial using the fundamental property $\cosh^2(x)-\sinh^2(x)=1$ (divide through by $\cosh^2(x)$).

Of course, partial fractions is the way to go if your teacher does not allow this form of an answer. In fact, the partial fraction decomposition and the identity $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$ would probably be a sorta neat way to prove that $\tanh^{-1}(x)=\frac{1}{2}(\log(1+x)-log(1-x))$. (Though another method of proving it: writing $y=\tanh(x)=\sinh(x)/\cosh(x)$, expanding into exponentials, and solving for $x$ in terms of $y$ isn't too difficult.)

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  • $\begingroup$ Indeed very nice :)! $\endgroup$ – J. W. Perry Jan 14 '14 at 4:34
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I think partial fractions may indeed be the "simplest" way, but you mentioned trigonometric substitution getting a bit hairy, and I figured I would give my best derivation. It is a useful skill. This would be advantageous speed wise only if you knew the integral of the secant function from memory.

Consider the more general integral, $$\int\frac{1}{a^2-v^2} \, dv.$$ If we let $$v=a\sin\theta,$$ then $$dv=a\cos \theta \, d \theta,$$ and $$a^2-v^2=a^2-a^2\sin^2 \theta=a^2\left( 1-\sin^2 \theta \right)=a^2 \cos^2 \theta.$$

We substitute to get $$ \begin{align*} \int \frac{1}{a^2-v^2} \, dv &= \int \frac{a \cos \theta \, d \theta}{a^2 \cos^2 \theta} \\ &=\frac{1}{a}\int\sec \theta \, d \theta \\ &=\frac{1}{a}\ln |\sec \theta + \tan \theta|+c. \end{align*} $$

For the back substitution, $$v=a \sin \theta \Rightarrow \sin \theta = \frac{v}{a}.$$

We form right a triangle with angle $\theta$, side $v$ opposite $\theta$, and hypotenuse $a$. Thus the side adjacent $\theta$ has length $\sqrt{a^2-v^2}$. Reading directly from the triangle, $$\frac{1}{a}\ln |\sec \theta + \tan \theta|+c=\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c.$$

We are done, but this certainly cleans up well (like in an integral table). $$ \begin{align*} &\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c \\ =& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a^2-v^2}} \right|+c \\ =& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a+v}\sqrt{a-v}} \right|+c \\ =&\frac{1}{a}\ln \left| \frac{\sqrt{a+v}}{\sqrt{a-v}} \right|+c \\ =&\frac{1}{a}\ln \left| \left( \frac{a+v}{a-v} \right)^{\frac{1}{2}} \right|+c \\ =&\frac{1}{2a}\ln \left| \frac{a+v}{a-v} \right|+c. \\ \end{align*} $$

Thus your particular integral is $$\int \frac{1}{4-v^2} \, dv=\frac{1}{4}\ln \left| \frac{a+v}{a-v} \right|+c.$$

It might not seem like the "easiest" way, but it is certainly doable, and in my opinion, fun.

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  • $\begingroup$ Hyperbolic functions are much more useful in this situation. $\endgroup$ – user18862 Jan 14 '14 at 4:22
  • $\begingroup$ It's definitely fun to do, but I didn't know the integral of sec at the time, and I was looking for a simpler, shorter method. $\endgroup$ – Korgan Rivera Jan 14 '14 at 18:16
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substitute $2z = v,$ and then $z = \sin \theta.$

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  • $\begingroup$ @anon told you I was asleep :( $\endgroup$ – Igor Rivin Jan 14 '14 at 1:47
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    $\begingroup$ this won't help. Use partial fraction as mentioned before $\endgroup$ – user44197 Jan 14 '14 at 1:47
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    $\begingroup$ @user44197 depends what you mean by "help". You get the integral of $\sec x,$ which is, admittedly. more difficult than the original, but might be known to the OP. $\endgroup$ – Igor Rivin Jan 14 '14 at 1:49
  • $\begingroup$ actually $\sec^2$ which is easier than $\sec$. In any case you end up with $\tan^2\theta$ and you have to write it in terms of $\sin\theta$ and finally $v$. $\endgroup$ – user44197 Jan 14 '14 at 1:57
  • $\begingroup$ @anon It looks like this substitution gives $\frac12\int\sec\theta d\theta$. $\endgroup$ – Mike Jan 14 '14 at 3:31

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