use any method to evaluate the integrals.\n int\frac{dx}{sqrt{169 - x^{2}}}\n int\frac{dx}{sqrt{169…
use any method to evaluate the integrals.\n int\frac{dx}{sqrt{169 - x^{2}}}\n int\frac{dx}{sqrt{169 - x^{2}}}=square
Answer
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:
$\arcsin(\frac{x}{13})+C$