Question

rewrite the given integral using this substitution.
\frac{5}{sqrt{2}}int_{0}^{\frac{5}{sqrt{2}}}\frac{dx}{sqrt{25 - x^{2}}}=int_{0}^{square}(square)d\theta
(type exact answers.)

Explanation:

Step1: Use substitution $x = 5\sin\theta$

If $x = 5\sin\theta$, then $dx=5\cos\theta d\theta$. Also, when $x = 0$, $0 = 5\sin\theta$ implies $\theta = 0$. When $x=\frac{5}{\sqrt{2}}$, $\frac{5}{\sqrt{2}}=5\sin\theta$, so $\sin\theta=\frac{1}{\sqrt{2}}$ and $\theta=\frac{\pi}{4}$. And $\sqrt{25 - x^{2}}=\sqrt{25-25\sin^{2}\theta}=5\cos\theta$.

Step2: Rewrite the integral

The integral $\int_{0}^{\frac{5}{\sqrt{2}}}\frac{dx}{\sqrt{25 - x^{2}}}$ becomes $\int_{0}^{\frac{\pi}{4}}\frac{5\cos\theta d\theta}{5\cos\theta}$. Simplifying the integrand, $\frac{5\cos\theta}{5\cos\theta}=1$. So the new - integral is $\int_{0}^{\frac{\pi}{4}}1\cdot d\theta$.

Answer:

$\int_{0}^{\frac{\pi}{4}}d\theta$