Let I=int_0^1 9/(3+x^2)^2\ dx. Using the substitution x=sqrt(3) tan(theta), show that I= sqrt(3) int_0^(pi/6) cos^2(theta)\ d theta. What is the exact value of I?

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Jan 28, 2018

${\int}_{0}^{1} \frac{9}{3 + {x}^{2}} ^ 2 \setminus \mathrm{dx} = \frac{\left(2 \pi + 3\right) \sqrt{3}}{24}$

Explanation:

$\setminus \setminus \setminus \setminus \setminus \setminus {\int}_{0}^{1} \frac{9}{3 + {x}^{2}} ^ 2 \setminus \mathrm{dx}$

Substitute $x = \sqrt{3} \tan \left(\theta\right)$ and $\mathrm{dx} = \sqrt{3} {\sec}^{2} \left(\theta\right) d \setminus \theta$:
$= {\int}_{\arctan} {\left(0\right)}^{\arctan} \left(\frac{1}{\sqrt{3}}\right) \frac{9}{3 + {\left(\sqrt{3} \tan \left(\theta\right)\right)}^{2}} ^ 2 \sqrt{3} {\sec}^{2} \left(\theta\right) \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} \frac{9 {\sec}^{2} \left(\theta\right)}{3 + 3 {\tan}^{2} \left(\theta\right)} ^ 2 \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\sec}^{2} \frac{\theta}{1 + {\tan}^{2} \left(\theta\right)} ^ 2 \setminus d \setminus \theta$

Using the fact that ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$:

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\sec}^{2} \frac{\theta}{\sec} ^ 4 \left(\theta\right) \setminus d \setminus \theta$

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} \frac{1}{\sec} ^ 2 \left(\theta\right) \setminus d \setminus \theta$

Since $\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$, we have

$= \sqrt{3} {\int}_{0}^{\frac{\pi}{6}} {\cos}^{2} \left(\theta\right) \setminus d \setminus \theta$

Using the fact that ${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$:

$= \frac{\sqrt{3}}{2} {\int}_{0}^{\frac{\pi}{6}} 1 + \cos \left(2 \theta\right) \setminus d \setminus \theta$

$= \frac{\sqrt{3}}{2} {\left[\theta + \sin \frac{2 \theta}{2}\right]}_{0}^{\frac{\pi}{6}}$

$= \frac{\sqrt{3}}{2} \left(\frac{\pi}{6} + \frac{1}{4}\right)$

$= \frac{\left(2 \pi + 3\right) \sqrt{3}}{24}$

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