Question

use any method to evaluate the integrals.
int\frac{dx}{sqrt{169 - x^{2}}}
int\frac{dx}{sqrt{169 - x^{2}}}=square

Explanation:

Step1: Recognize the integral form

We know that the integral $\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\arcsin(\frac{x}{a})+C$, where $a > 0$. Here $a^{2}=169$, so $a = 13$.

Step2: Apply the formula

Substitute $a = 13$ into the formula. Then $\int\frac{dx}{\sqrt{169 - x^{2}}}=\int\frac{dx}{\sqrt{13^{2}-x^{2}}}=\arcsin(\frac{x}{13})+C$.

Answer:

$\arcsin(\frac{x}{13})+C$