Question

use any method to evaluate the integral
$int \frac{242 dw}{w^{2}sqrt{121 - w^{2}}}$
$int \frac{242 dw}{w^{2}sqrt{121 - w^{2}}} = square$
(type an exact answer, using radicals as

Explanation:

Step1: Trigonometric substitution

Let $w = 11\sin\theta$, so $dw = 11\cos\theta d\theta$.
Also, $\sqrt{121-w^2} = \sqrt{121-121\sin^2\theta} = 11\cos\theta$.

Step2: Substitute into integral

$$\begin{align*} \int \frac{242}{w^2\sqrt{121-w^2}} dw &= \int \frac{242}{(11\sin\theta)^2 \cdot 11\cos\theta} \cdot 11\cos\theta d\theta\\ &= \int \frac{242}{121\sin^2\theta} d\theta\\ &= 2\int \csc^2\theta d\theta \end{align*}$$

Step3: Evaluate the integral

$\int \csc^2\theta d\theta = -\cot\theta + C$, so:
$2\int \csc^2\theta d\theta = -2\cot\theta + C$

Step4: Convert back to $w$

Since $w=11\sin\theta$, $\sin\theta = \frac{w}{11}$, so $\cot\theta = \frac{\sqrt{121-w^2}}{w}$.
Substitute back: $-2\cdot\frac{\sqrt{121-w^2}}{w} + C$

Answer:

$-\frac{2\sqrt{121-w^2}}{w} + C$