Trig Subs issues

Question

Problem:

Use a trigonometric substitution to find

$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$

Give both an exact answer (involving $\pi$ and a square root) and a decimal estimate to 3 significant digits.

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Here is the work I have so far.

$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$

$$1 + \tan^2\Theta = \sec^2\Theta$$

$$\sec^2\Theta - 1 =\ tan^2\Theta$$

$$x = \sqrt{3}\sec\Theta$$

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$$\sqrt{(\sqrt{3}\sec\Theta)^2 - 3}$$

$$\sqrt{3(1+\tan^2\Theta)-3}$$

$$\sqrt{3\tan^2\Theta}$$

$$\sqrt{3} \ tan\Theta$$

$$I = \int_\sqrt{3}^2{\frac{\sqrt{3}\tan\Theta}{\sqrt{3}\sec\Theta}}\;dx$$

$$\frac{dx}{d\Theta} = (\sqrt{3}\sec\Theta)'$$

$$dx = \sqrt{3}\tan\Theta\sec \Theta d\Theta$$

I feel like around here I may have messed up and I'm not sure if I should switch the limits of integration to $\Theta$ numbers of $0$ to $\frac{\pi}{6}$ yet or not. Any help is appreciated =) Thanks.

Answer 1

You did well.

Swith the limit when you replace $dx$ by $d\Theta$.

\begin{align} I&= \int_0^{\frac{\pi}6} \frac{\tan \Theta}{\sec \Theta} \sqrt{3} \tan \Theta \sec \Theta \, d\Theta \\ &= \sqrt{3} \int_0^{\frac{\pi}{6}} \tan^2\Theta\, d\Theta\\ &= \sqrt{3} \int_0^{\frac{\pi}6} \sec^2 \Theta -1 \,d\Theta\\ &= \sqrt{3} \left( \tan \Theta - \Theta\right) \mid_0^\frac{\pi}{6} \\ &= \sqrt{3} \left( \frac1{\sqrt3}-\frac{\pi}6\right)\\ &= 1 - \frac{\pi\sqrt{3}}{6} \end{align}

Answer 2

I give a method other than trig sub. Let $t=\sqrt{x^2-3}$, then $x^2=t^2+3$. Note $dt=\dfrac{x}{\sqrt{x^2-3}}dx$. Hence, we have $$\int^2_{\sqrt{3}}\dfrac{\sqrt{x^2-3}}{x}~dx=\int^{1}_0\dfrac{t^2}{t^2+3}~dt=1-\int^1_0\dfrac{3}{t^2+3}~dt=1-\left[\dfrac{3}{\sqrt{3}}\arctan\left(\dfrac{t}{\sqrt{3}}\right)\right]^1_0=1-\dfrac{\pi}{2\sqrt{3}}$$