Questions: [ int frac36 d ww^2 sqrt9-w^2 ]

Solution Steps

To evaluate the integral \(\int \frac{36 dw}{w^{2} \sqrt{9-w^{2}}}\), we can use a trigonometric substitution. Since the integrand involves a square root of the form \(\sqrt{a^2 - w^2}\), we can use the substitution \(w = 3 \sin \theta\). This substitution will simplify the square root and allow us to integrate with respect to \(\theta\).

The given integral is:

\[ \int \frac{36 \, dw}{w^{2} \sqrt{9-w^{2}}} \]

This integral involves a square root of the form \(\sqrt{a^2 - w^2}\), which suggests a trigonometric substitution. Specifically, we can use the substitution \(w = 3 \sin \theta\), where \(a = 3\).

Using the substitution \(w = 3 \sin \theta\), we have:

\[ dw = 3 \cos \theta \, d\theta \]

Substituting these into the integral, we get:

\[ \int \frac{36 \cdot 3 \cos \theta \, d\theta}{(3 \sin \theta)^2 \sqrt{9 - (3 \sin \theta)^2}} \]

Simplifying the expression inside the integral:

\[ = \int \frac{108 \cos \theta \, d\theta}{9 \sin^2 \theta \sqrt{9 - 9 \sin^2 \theta}} \]

The term \(\sqrt{9 - 9 \sin^2 \theta}\) simplifies to:

\[ \sqrt{9(1 - \sin^2 \theta)} = \sqrt{9 \cos^2 \theta} = 3 \cos \theta \]

Substituting this back, the integral becomes:

\[ = \int \frac{108 \cos \theta \, d\theta}{9 \sin^2 \theta \cdot 3 \cos \theta} \]

Cancel out the common terms:

\[ = \int \frac{108 \, d\theta}{27 \sin^2 \theta} \]

\[ = \int \frac{4 \, d\theta}{\sin^2 \theta} \]

Recognizing that \(\frac{1}{\sin^2 \theta} = \csc^2 \theta\), the integral becomes:

\[ = 4 \int \csc^2 \theta \, d\theta \]

The integral of \(\csc^2 \theta\) is \(-\cot \theta\):

\[ = 4(-\cot \theta) + C \]

\[ = -4 \cot \theta + C \]

Recall the substitution \(w = 3 \sin \theta\), which implies \(\sin \theta = \frac{w}{3}\). Therefore, \(\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} = \frac{\sqrt{1 - \left(\frac{w}{3}\right)^2}}{\frac{w}{3}}\).

Simplifying:

\[ \cot \theta = \frac{\sqrt{\frac{9 - w^2}{9}}}{\frac{w}{3}} = \frac{\sqrt{9 - w^2}}{w} \]

Substituting back, we have:

\[ -4 \cot \theta = -4 \cdot \frac{\sqrt{9 - w^2}}{w} \]

Final Answer