Troubleshooting a difficult integral

[{???}]

Hello everyone,

I have a maths question (for a change). In summary, I would like to reconcile the following two integrals:
Integral A: https://www.wolframalpha.com/input/?i=integrate+(a^2tan^2theta)/(a-b+cos+theta)+dtheta
$$\int\frac{x^2\,dx}{\sqrt{x^2+a^2}(\sqrt{x^2+a^2}-b)} =x +b\ln(\sqrt{a^2+x^2}+x) -\sqrt{a^2-b^2}\tan^{-1}\frac{bx}{\sqrt{a^2-b^2}\sqrt{a^2+x^2}} -\sqrt{a^2-b^2}\tan^{-1}\frac{x}{\sqrt{a^2-b^2}} +C.$$
Integral B: https://www.wolframalpha.com/input/?i=integrate+x^2/(sqrt(x^2+a^2)*(sqrt(x^2+a^2)-b))
$$\int\frac{a^2\tan^2\theta\,d\theta}{a-b\cos\theta} = a\tan\theta +b\ln\frac{\cos\tfrac{\theta}{2}+\sin\tfrac{\theta}{2}}{\cos\tfrac{\theta}{2}-\sin\tfrac{\theta}{2}} -2\sqrt{a^2-b^2}\tan^{-1}\left(\sqrt\frac{a+b}{a-b}\tan\frac{\theta}{2}\right) +C.$$
Note that Integral B is integral A with the substitution $x=a\tan\theta$. Thus terms 1 and 2 in Integral A are the same as terms 1 and 2 in Integral B (term #2 requires some trigonometric manipulation and borrows a factor $\ln a$ from the integration constant).

My question is about the remaining terms. Namely, how do I show that
$$\tan^{-1}\frac{bx}{\sqrt{a^2-b^2}\sqrt{a^2+x^2}} +\tan^{-1}\frac{x}{\sqrt{a^2-b^2}} +C= 2\tan^{-1}\left(\sqrt\frac{a+b}{a-b}\tan\left[\frac{1}{2}\tan^{-1}\frac{x}{a}\right]\right) +C'?$$
I allow $C\neq C'$ in case this turns out to be necessary. I have tried using trigonometric identities (e.g. $\tan(A+B),\tan(\theta/2)$), and also computing the derivatives of both expressions to check that they are exactly equal, but this doesn't seem to work. I know the following:
1. Integral A is correct (checking the derivative of the result is equal to the integrand).
2. I can reproduce Integral B step-by-step (will supply proof if necessary).

Can anyone please shed some light on this problem? I would certainly appreciate it!

Cheerio,
QM

 

[eys_physics]

Have you tried to use the addition formula of $\tan^{-1}$? That is,
$$\tan^{-1}(u) + \tan^{-1}(v) =\tan^{-1}\left(\frac{u + v}{1 - uv}\right)\quad (mod\: \pi)$$

 

[Fred Wright]

Your $\theta$ integral is wrong. You have,$$

\int \frac{x^2dx}{(x^2 +a^2) -b\sqrt((x^2 +a^2))}
$$
Substitute $x=a\tan(\theta)$$$
dx=\frac{ad\theta}{\cos^2(\theta)}
$$
Therefore your $\theta$ integral should be,
$$
\int \frac{\tan^2(\theta)\cos^4(\theta)d\theta}{a-b\cos(\theta)}
$$

 

[{???}]

Have you tried to use the addition formula of $\tan^{-1}$? That is,
$$\tan^{-1}(u) + \tan^{-1}(v) =\tan^{-1}\left(\frac{u + v}{1 - uv}\right)\quad (mod\: \pi)$$

Yes:

I have tried using trigonometric identities (e.g. $\tan(A+B),\tan(\theta/2)$), and also computing the derivatives of both expressions to check that they are exactly equal...

The formula you quote is just the inverse of the identity $\tan(A+B)=(\tan A+\tan B)/(1-\tan A\tan B)$. While this seems an extremely obvious way to proceed, the expressions I get just become messier and do not begin to resemble the other side of the tentative equation I wish to prove.

Your $\theta$ integral is wrong.

$$
dx=\frac{ad\theta}{\cos^2(\theta)}
$$
Therefore your $\theta$ integral should be,
$$
\int \frac{\tan^2(\theta)\cos^4(\theta)d\theta}{a-b\cos(\theta)}
$$

No, your integral is wrong. You have multiplied by $\cos^2\theta$ in the final line instead of dividing by it.

My calculation is correct and I can prove it. Integral A:
$$\int\frac{x^2\,dx}{\sqrt{x^2+a^2}(\sqrt{x^2+a^2}-b)}$$
Substitute $x=a\tan\theta,dx=a\sec^2\theta\,d\theta$:
$$\int\frac{a^3\tan^2\theta\sec^2\theta\,d\theta}{\sqrt{a^2(1+\tan^2\theta)}(\sqrt{a^2(1+\tan^2\theta)}-b)}$$
$$=\int\frac{a^3\tan^2\theta\sec^2\theta\,d\theta}{a\sec\theta(a\sec\theta-b)}$$
$$=\int\frac{a^2\tan^2\theta\sec\theta\,d\theta}{a\sec\theta-b}$$
$$=\int\frac{a^2\tan^2\theta\,d\theta}{a-b\cos\theta}$$
If you disagree with any line of my arithmetic, please quote me there specifically.

I am going to try to reverse-engineer WA's answer by differentiating the terms in question, substituting as I would to solve the integral, and identifying which partial fraction is supposed to yield this integral.

 

[{???}]

FYI I solved my problem on my own.

While my solution is perfectly valid, it uses a Weierstrass substitution which is actually not needed for the integral (although it is quite surprising that it is not necessary!). As luck would have it, the integral can be subdued without the sneaky $\theta=2\phi$ substitution, and by doing so we reproduce the WA result.

Moderators, please feel free to close this thread.

 

[Fred Wright]

Yes:

The formula you quote is just the inverse of the identity $\tan(A+B)=(\tan A+\tan B)/(1-\tan A\tan B)$. While this seems an extremely obvious way to proceed, the expressions I get just become messier and do not begin to resemble the other side of the tentative equation I wish to prove.
No, your integral is wrong. You have multiplied by $\cos^2\theta$ in the final line instead of dividing by it.

My calculation is correct and I can prove it. Integral A:
$$\int\frac{x^2\,dx}{\sqrt{x^2+a^2}(\sqrt{x^2+a^2}-b)}$$
Substitute $x=a\tan\theta,dx=a\sec^2\theta\,d\theta$:
$$\int\frac{a^3\tan^2\theta\sec^2\theta\,d\theta}{\sqrt{a^2(1+\tan^2\theta)}(\sqrt{a^2(1+\tan^2\theta)}-b)}$$
$$=\int\frac{a^3\tan^2\theta\sec^2\theta\,d\theta}{a\sec\theta(a\sec\theta-b)}$$
$$=\int\frac{a^2\tan^2\theta\sec\theta\,d\theta}{a\sec\theta-b}$$
$$=\int\frac{a^2\tan^2\theta\,d\theta}{a-b\cos\theta}$$
If you disagree with any line of my arithmetic, please quote me there specifically.

I am going to try to reverse-engineer WA's answer by differentiating the terms in question, substituting as I would to solve the integral, and identifying which partial fraction is supposed to yield this integral.

I apologize. Your right. I'm wrong. It must be my dyslexia.

 

[{???}]

I apologize. Your right. I'm wrong. It must be my dyslexia.

No worries - It's a very tricky integral even (or perhaps especially?) without the Weierstrass substitution. :)

I have noticed that integrals arising from a physical scenario (rather than simply composed for mathematics) tend to have "nicer" answers. Without giving away too much, the integrals I discuss do indeed arise from a simple, if challenging, physics exercise. I just wanted to make sure they can be done and that the answer I get is correct.

Cheers,
QM