evaluate the following integral using trigonometric substitution.\n int_{0}^{\frac{5}{sqrt{2}}}…

Explanation:

Step1: Choose substitution

For the integral $\int\frac{dx}{\sqrt{25 - x^{2}}}$, we use the substitution $x = 5\sin\theta$. Then $dx=5\cos\theta d\theta$. Also, $\sqrt{25 - x^{2}}=\sqrt{25-25\sin^{2}\theta}=5\cos\theta$.

Step2: Change limits

When $x = 0$, $0 = 5\sin\theta$, so $\theta=0$. When $x=\frac{5}{\sqrt{2}}$, $\frac{5}{\sqrt{2}}=5\sin\theta$, then $\sin\theta=\frac{1}{\sqrt{2}}$, so $\theta=\frac{\pi}{4}$.

Step3: Rewrite integral

The integral $\int_{0}^{\frac{5}{\sqrt{2}}}\frac{dx}{\sqrt{25 - x^{2}}}$ becomes $\int_{0}^{\frac{\pi}{4}}\frac{5\cos\theta d\theta}{5\cos\theta}=\int_{0}^{\frac{\pi}{4}}d\theta$.

Step4: Evaluate integral

$\int_{0}^{\frac{\pi}{4}}d\theta=\left[\theta\right]_{0}^{\frac{\pi}{4}}=\frac{\pi}{4}-0=\frac{\pi}{4}$.

Answer:

B. $x = 5\sin\theta$ $\frac{\pi}{4}$