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$$\int \dfrac{dx}{x\sqrt{x^4-1}}$$

I need to solve this integration. I solved and got $\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$, however the answer given in my textbook is $\dfrac12\sec^{-1}(x^2) + C$

How can I prove that both quantities are equal? Is there something wrong with my answer?

EDIT:

Here's my work: $$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$

Let $x^4 - 1 = t^2$ $$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$

$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$

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    $\begingroup$ Post your solution as well please. $\endgroup$
    – youthdoo
    Mar 14 at 13:17
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    $\begingroup$ Hint: if $\tan y=\sqrt{x^4-1}$ then $\sec y=\pm\sqrt{1+\tan^2y}=\cdots$. $\endgroup$
    – J.G.
    Mar 14 at 13:22

4 Answers 4

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I tell my students that inverse trig functions are angles. So if you write

$$\tan^{-1}\sqrt{x^4-1} = \theta,$$

then

$$\tan\theta = \sqrt{x^4-1}.$$

A right triangle that tells this story has $\theta$ as one angle, $\sqrt{x^4-1}$ as the opposite side and $1$ as the adjacent side. Using Pythagorean theorem we can work out the length $c$ of the hypotenuse:

$$(\sqrt{x^4-1})^2+1^2 = c^2$$

which shows that $c=x^2$.

So $\sec \theta = x^2/1$, that is $\sec^{-1}(x^2) = \theta = \tan^{-1}( \sqrt{x^4-1}).$

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    $\begingroup$ It should be added that if $\tan^{-1} \sqrt{x^4-1} = \theta$ then $0 \le \theta < \pi/2$, since $\sqrt{x^4-1} \ge 0$. That is needed to justify making a right triangle with angle $\theta$. $\endgroup$ Mar 14 at 15:34
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The answer was already given by B. Goddard, here I try to give a visual answer:

enter image description here

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    $\begingroup$ Awesome, I didn't expect this. Thanks a ton! $\endgroup$
    – user983206
    Mar 14 at 15:33
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I think your problem is already solved with B. Goddard's answer, Here's an easy way to solve the given integral.


We have, $$\int\dfrac{dx}{x\sqrt{x^4-1}}$$

Let $x^2 = \sec(\theta) \implies 2x \, dx = \sec(\theta) \tan(\theta) \, d\theta$

Which further implies that, $dx = \dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{2}\, d\theta$

After substitution, the given integral changes to: $$\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\sec^2(\theta) - 1}}\, d\theta$$

$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\tan^2(\theta)}}\, d\theta$$

$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\tan(\theta)}\, d\theta$$

$$=\dfrac12\int d\theta$$

$$=\dfrac12\theta + C$$

$$=\boxed{\dfrac12\sec^{-1}(x^2) + C}$$

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    $\begingroup$ Woah! Nice substitution. $\endgroup$
    – user983206
    Mar 14 at 15:31
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There is nothing wrong with your answer. The antiderivative you found is correct.

Your book's answer is also correct. The thing is that

$$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})=\frac{1}{2}\sec^{-1}(x^2)$$

If you don't believe me, see their graphs. The graphs of $\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})$ and $\frac{1}{2}\sec^{-1}(x^2)$ are exactly the same.

Edit:

In order to prove that the quantities are indeed equal see @Zaragosa's answer.


PS: Whenever in doubt about calculus, use the derivative-calculator or the integral-calculator.

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