Calculating $\int \frac{x}{\sqrt{x^{2}-36}} \, {\rm d} x$ [duplicate]

Question

I wish to calculate

$$\int \frac{x}{\sqrt{x^{2}-36}} \, {\rm d} x$$

I use a substitution $x=6\sec\theta$ and get $\int \csc\theta d\theta$ as intermediate.. Then, after some calculation

$$-\ln\left|\frac{x}{\sqrt{x^2-36}}+\frac{6}{x^2-36}\right|$$

Is it correct?

If you substitute $x=6\sec\theta$, the integral becomes $36\int\sec\theta\tan^2\theta\,d\theta$. How do you get $\csc\theta$ as the integrand? — 2'5 9'2, Mar 4, 2021 at 3:13
quora.com/How-do-you-integrate-sqrt-x-2-a-2-with-respect-to-x — P. J., Mar 4, 2021 at 5:26
Note that there's also a geometric interpretation of this integral. If we turn this into a definite integral from $a$ to $b$, you can think of it as the area under the arc of a circle with radius $6$ between $a$ and $b$, which can be decomposed into a fan-shape plus two triangles. That should give you an inkling for what the result might look like. — Elliot Yu, Mar 4, 2021 at 5:27
It makes more sense to substitute $x=6\sin\theta$ (where $\pi/2 < \theta < \pi/2$). The integrals ends up looking much nicer. — Joe, Mar 4, 2021 at 5:40

Nikola Alfredi · Accepted Answer · 2021-03-04 06:20:12Z

Go easy on it. Take

$\displaystyle x^2 - 36 = y \ \implies 2x \ \mathrm {d}x = \mathrm {d}y$

$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1/2} \ \mathrm {d}y = \sqrt {y} + c .$$

Substitute the value of $y$ and you get :

$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \sqrt {x^2 -36} + c .$$

But as mentioned by @integration brainstorm : A typo in the Question, using same substitution :

$$\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1} \ \mathrm {d}y = \frac {\ln {y}}{2} + c . $$

Thus $\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {\ln (x^2 -36)}{2} + c . $

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