evaluate the following integral.\n int_{0}^{5/sqrt{2}} \frac{dx}{sqrt{25 - x^{2}}}\n int_{0}^{5/sqrt{2}}…

evaluate the following integral.\n int_{0}^{5/sqrt{2}} \frac{dx}{sqrt{25 - x^{2}}}\n int_{0}^{5/sqrt{2}} \frac{dx}{sqrt{25 - x^{2}}}=square (type an exact answer, using (pi) as needed.)

evaluate the following integral.\n int_{0}^{5/sqrt{2}} \frac{dx}{sqrt{25 - x^{2}}}\n int_{0}^{5/sqrt{2}} \frac{dx}{sqrt{25 - x^{2}}}=square (type an exact answer, using (pi) as needed.)

Answer

Explanation:

Step1: Recall integral formula

The integral $\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\arcsin(\frac{x}{a})+C$, where $a > 0$. Here $a = 5$.

Step2: Apply the fundamental theorem of calculus

$\int_{0}^{\frac{5}{\sqrt{2}}}\frac{dx}{\sqrt{25 - x^{2}}}=\left[\arcsin(\frac{x}{5})\right]_{0}^{\frac{5}{\sqrt{2}}}$

Step3: Evaluate the definite - integral

$\arcsin(\frac{\frac{5}{\sqrt{2}}}{5})-\arcsin(0)=\arcsin(\frac{1}{\sqrt{2}})-0$ Since $\arcsin(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$, the value of the integral is $\frac{\pi}{4}$.

Answer:

$\frac{\pi}{4}$