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I have been trying to solve:

$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$

I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$

So then I have:

$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$

$$\int \frac{\sqrt{9(\sec^2 \theta - 1)}}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$

$$ \int \frac{3\tan \theta}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$

$$ = \int \frac{9 \tan ^2 \theta}{27 \sec ^2 \theta} d \theta$$

$$ = \int \frac{\tan ^2 \theta}{3 \sec ^2 \theta} d \theta$$

$$ \int \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{3} d \theta$$

$$ \int \frac{\sin^2 \theta}{3} d \theta$$

$$\frac{1}{3} \int \sin^2 \theta d \theta$$

Am I on the right track?

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  • $\begingroup$ Yes, you. However, I think a hyperbolic substitution would be a bit shorter. $\endgroup$ – Bernard Feb 3 at 16:46
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    $\begingroup$ On the right track except for some missing $=$ and $d\theta$. $\endgroup$ – Bernard Massé Feb 3 at 17:06
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Yes, your solution so far is correct. Now, to integrate $\int \sin^2{\theta}\,d\theta$, use the half-angle formula for the sine function:

$$ \sin^2{\theta}=\frac{1-\cos(2\theta)}{2}. $$

Also, a bit later, you're going to need this formula:

$$ \sin{(2\theta)}=2\sin{\theta}\cos{\theta}. $$

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$$I=\int\frac{\sqrt{x^2-9}}{x^3}dx=\int\frac{3\sqrt{(x/3)^2-1}}{x^3}dx$$ and we know that $\cosh^2\theta-1=\sinh^2\theta$ $$x=3\cosh(y)$$ $$dx=3\sinh(y)dy$$ $$I=9\int\frac{\sqrt{\cosh^2(y)-1}}{27\cosh^3(y)}\sinh(y)dy=\frac13\int\frac{\sinh^2(y)}{\cosh^3(y)}dy=\frac13\int\text{sech}(y)-\text{sech}^3(y)dy$$ a reduction formula can then be used

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