I'm trying to integrate the following:

$$\int \frac {dx}{x\sqrt{x^2-49}}\,$$

using the substitution $x=7\cosh(t)$

This is as far as I've gotten:

$\int \frac {dx}{x\sqrt{x^2-49}}\,$ = $\int \frac {7\sinh(t)dt}{7\cosh(t)7\sinh(t)}\,$ = $\int \frac {dt}{7\cosh(t)}\,$ = $\int \frac {\cosh(t)dt}{7\cosh^2(t)}\,$ = $\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\,$

Let $u=\sinh(t)$, $du=\cosh(t)dt$

$$\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\, =\int \frac {du}{7(1+u^2)}\,$$ $$=\frac {1}{7}\arctan(u)+C=\frac{1}{7} \arctan(\sinh(t))+C$$

This is as far as I have been able to get. Somehow from here I need to get to

$$-\frac{1}{7} \arctan(\frac{7}{\sqrt{x^2-49}})+C$$

Can someone please show me how to finish this integration problem off? I would appreciate it so so much.

  • 3
    $\begingroup$ Another change: consider $t=\sqrt{x^2-49}$. $\endgroup$
    – A.Γ.
    May 18 '20 at 4:54
  • $\begingroup$ math.stackexchange.com/questions/1606590/… $\endgroup$ May 18 '20 at 4:58
  • $\begingroup$ Yet another substitution: $t = 7 \sec \theta$. I think the reason that OP uses hyperbolic functions is because the question says so. $\endgroup$
    – Toby Mak
    May 18 '20 at 4:59


$$\frac{1}{7} \arctan(\sinh t) =\frac{1}{7} \arctan\sqrt{\cosh^2t -1}\\ =\frac{1}{7} \arctan\sqrt{\frac{x^2}{49}-1} =\frac{1}{7} \arctan\frac{\sqrt{x^2-49}}7\\ = \frac{1}{7} \text{arccot } \frac7{\sqrt{x^2-49}} = \frac{1}{7}(\frac\pi2- \text{arctan} \frac7{\sqrt{x^2-49}})\\ =- \frac{1}{7}\text{arctan} \frac7{\sqrt{x^2-49}}+C $$


My advice is to make the substitution $x=7 \sec u$ in the original quesion. Then all that remains is to integrate $\frac{1}{7}.$



$\cosh^2 t - \sinh^2 t =1$ so $\sinh t = \sqrt{\cosh^2 t- 1}$.

Then refer back to the substitution $x = 7 \cosh t$.


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