Questions: Evaluate ∫ from -8√3 to 8√3 of dx / (64 + x^2)

Solution Steps

We need to evaluate the definite integral

\[ \int_{-8 \sqrt{3}}^{8 \sqrt{3}} \frac{\mathrm{dx}}{64+\mathrm{x}^{2}}. \]

This integral can be recognized as a standard form of the integral of the function \(\frac{1}{a^2 + x^2}\), where \(a = 8\).

The antiderivative of \(\frac{1}{64 + x^2}\) is given by

\[ \frac{1}{8} \arctan\left(\frac{x}{8}\right). \]

We evaluate the antiderivative at the upper and lower bounds:

• For \(x = 8\sqrt{3}\):

\[ F(8\sqrt{3}) = \frac{1}{8} \arctan\left(\frac{8\sqrt{3}}{8}\right) = \frac{1}{8} \arctan(\sqrt{3}) = \frac{1}{8} \cdot \frac{\pi}{3} = \frac{\pi}{24}. \]

• For \(x = -8\sqrt{3}\):

\[ F(-8\sqrt{3}) = \frac{1}{8} \arctan\left(\frac{-8\sqrt{3}}{8}\right) = \frac{1}{8} \arctan(-\sqrt{3}) = \frac{1}{8} \cdot \left(-\frac{\pi}{3}\right) = -\frac{\pi}{24}. \]

Now, we find the definite integral by subtracting the values of the antiderivative at the bounds:

\[ \int_{-8 \sqrt{3}}^{8 \sqrt{3}} \frac{\mathrm{dx}}{64+\mathrm{x}^{2}} = F(8\sqrt{3}) - F(-8\sqrt{3}) = \frac{\pi}{24} - \left(-\frac{\pi}{24}\right) = \frac{\pi}{24} + \frac{\pi}{24} = \frac{2\pi}{24} = \frac{\pi}{12}. \]

Final Answer