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}^{\frac{pi}{7}}(square)d\theta
(type exact answers.)

Explanation:

Step1: Use trig - 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 new 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 $\theta=\frac{\pi}{4}$. But we are given the upper - limit as $\frac{\pi}{7}$, so we use the substitution with the given upper - limit.

Step3: Rewrite the integral

Substitute $x = 5\sin\theta$, $dx = 5\cos\theta d\theta$ and $\sqrt{25 - x^{2}}=5\cos\theta$ into the integral:
\[

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

\]

Answer:

$\int_{0}^{\frac{\pi}{7}}1d\theta$