<s>[INST] Integration problem yields two solutions. But are they the same? I have an integration problem: $\int\frac{1}{x\sqrt {x^{12}-1}}dx$.
My attempt was to let $u=\sqrt {x^{12}-1}$. Solving it yields:
$$\frac{1}{6}\int\frac{1}{u^2+1}du$$
$$=\frac{1}{6}\tan^{-1}(u)+C$$
$$=\frac{1}{6}\tan^{-1}\sqrt {x^{12}-1}+C$$
However, my TA substituted $u=\frac{1}{x}$, and subsequently $v=u^6$. Solving it yields:
$$-\frac{1}{6}\int\frac{1}{\sqrt {1-v^2}}dv$$
$$=-\frac{1}{6}\sin^{-1}(v)+C$$
$$=-\frac{1}{6}\sin^{-1}\frac{1}{x^6}+C$$
I do not understand how the two answers are different but mathematically are correct. When I tried to draw a right-angled triangle using the first solution, I would have:
opposite = $\sqrt{x^{12}-1}$
adjacent = $1$
hypotenuse = $x^6$
So from the above, $\sin\theta=\frac{\sqrt{x^{12}-1}}{x^6}$ and $\theta=\arcsin\frac{\sqrt{x^{12}-1}}{x^6}$, which is not the same as $-\arcsin\frac{1}{x^6}$
May I know what is wrong with my deduction above?
Note: I also tried to differentiate both solutions. The first solution will yield back to the integrand. But the second solution does not. But strangely, if I use any arbitrary limits on the two solutions, both give me the same answer.
Thank you in advance.
 [/INST] According to your consideration,
$\cos\theta = \frac{1}{x^6} \Rightarrow \theta = \cos^{-1}(\frac1{x^6})$
And $\cos^{-1}x = \frac\pi2-\sin^{-1}x$.
So, $\boxed{\theta = \frac\pi2 - \sin^{-1}\left(\frac{1}{x^6}\right)}$.
Now, $\frac{1}{6}\cdot\frac\pi2$ is just a constant and you have the required result as $\boxed{- \frac16\sin^{-1}\left(\frac{1}{x^6}\right) +k}$
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