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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
[tex]
\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.
[/tex]
Integral B: https://www.wolframalpha.com/input/?i=integrate+x^2/(sqrt(x^2+a^2)*(sqrt(x^2+a^2)-b))
[tex]
\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.
[/tex]
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
[tex]
\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'?
[/tex]
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
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
[tex]
\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.
[/tex]
Integral B: https://www.wolframalpha.com/input/?i=integrate+x^2/(sqrt(x^2+a^2)*(sqrt(x^2+a^2)-b))
[tex]
\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.
[/tex]
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
[tex]
\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'?
[/tex]
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