Question

evaluate $int_{-5sqrt{3}/3}^{5sqrt{3}/3}\frac{dx}{25 + x^{2}}$. $int_{-5sqrt{3}/3}^{5sqrt{3}/3}\frac{dx}{25 + x^{2}}=square$ (type an exact answer, using $pi$ and radicals as needed)

Explanation:

Step1: Recall integral formula

Recall that $\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\arctan(\frac{x}{a})+C$. Here $a = 5$. So $\int\frac{dx}{25 + x^{2}}=\frac{1}{5}\arctan(\frac{x}{5})+C$.

Step2: Apply fundamental theorem of calculus

Let $F(x)=\frac{1}{5}\arctan(\frac{x}{5})$. Then $\int_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}\frac{dx}{25 + x^{2}}=F(\frac{5\sqrt{3}}{3})-F(-\frac{5\sqrt{3}}{3})$.
\[

$$\begin{align*} F(\frac{5\sqrt{3}}{3})&=\frac{1}{5}\arctan(\frac{\frac{5\sqrt{3}}{3}}{5})=\frac{1}{5}\arctan(\frac{\sqrt{3}}{3})\\ F(-\frac{5\sqrt{3}}{3})&=\frac{1}{5}\arctan(\frac{-\frac{5\sqrt{3}}{3}}{5})=\frac{1}{5}\arctan(-\frac{\sqrt{3}}{3}) \end{align*}$$

\]
Since $\arctan(-x)=-\arctan(x)$, we have:
\[

$$\begin{align*} F(\frac{5\sqrt{3}}{3})-F(-\frac{5\sqrt{3}}{3})&=\frac{1}{5}\arctan(\frac{\sqrt{3}}{3})-\frac{1}{5}\arctan(-\frac{\sqrt{3}}{3})\\ &=\frac{1}{5}\arctan(\frac{\sqrt{3}}{3})+\frac{1}{5}\arctan(\frac{\sqrt{3}}{3})\\ &=\frac{2}{5}\arctan(\frac{\sqrt{3}}{3}) \end{align*}$$

\]
And $\arctan(\frac{\sqrt{3}}{3})=\frac{\pi}{6}$, so $\frac{2}{5}\arctan(\frac{\sqrt{3}}{3})=\frac{2}{5}\times\frac{\pi}{6}=\frac{\pi}{15}$.

Answer:

$\frac{\pi}{15}$