Question

rewrite the given integral using this substitution.
$\frac{5}{sqrt{2}}int_{0}\frac{dx}{sqrt{25 - x^{2}}}=int_{0}^{\frac{pi}{7}}(\textcircled{1})d\theta$
(type exact answers.)

Explanation:

Step1: Set up the substitution

Let $x = 5\sin\theta$, then $dx=5\cos\theta d\theta$. Also, $\sqrt{25 - x^{2}}=\sqrt{25-25\sin^{2}\theta}=5\cos\theta$.

Step2: Find the new limits of integration

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 $\theta=\frac{\pi}{4}$. But we are given the upper - limit as $\frac{\pi}{7}$ in the problem.
Substitute $x$ and $dx$ into the integral:
\[

$$\begin{align*} \int_{0}^{\frac{5}{\sqrt{2}}}\frac{dx}{\sqrt{25 - x^{2}}}&=\int_{0}^{\frac{\pi}{7}}\frac{5\cos\theta d\theta}{5\cos\theta}\\ \end{align*}$$

\]

Step3: Simplify the integrand

The $5\cos\theta$ terms in the numerator and denominator cancel out. So the integrand is $1$.

Answer:

$1$