Questions: Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) [ int fracduu sqrt2-u^2 ]

$Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) [ int fracduu sqrt2-u^2 ]$

Transcript text: Evaluate the integral. (Remember to use absolute values where appropriate. Use $C$ f \[ \int \frac{d u}{u \sqrt{2-u^{2}}} \]

Solution

Solution Steps

Step 1: Identify the Type of Integral

The given integral is:

\[ \int \frac{d u}{u \sqrt{2-u^{2}}} \]

This integral involves a square root of a quadratic expression, which suggests that a trigonometric substitution might be useful.

Step 2: Choose an Appropriate Substitution

To simplify the integral, we can use the substitution $ u = \sqrt{2} \sin \theta $. This implies that $ du = \sqrt{2} \cos \theta \, d\theta $ and $ \sqrt{2 - u^2} = \sqrt{2} \cos \theta $.

Step 3: Substitute and Simplify

Substitute $ u = \sqrt{2} \sin \theta $ into the integral:

\[ \int \frac{\sqrt{2} \cos \theta \, d\theta}{\sqrt{2} \sin \theta \cdot \sqrt{2} \cos \theta} = \int \frac{\sqrt{2} \cos \theta \, d\theta}{2 \sin \theta \cos \theta} \]

This simplifies to:

\[ \int \frac{d\theta}{\sin \theta} \]

Step 4: Solve the Simplified Integral

The integral $\int \frac{d\theta}{\sin \theta}$ is a standard integral, which is equal to $-\ln |\csc \theta + \cot \theta| + C$.

Step 5: Back-Substitute to Original Variable

Recall that $ u = \sqrt{2} \sin \theta $, so $\sin \theta = \frac{u}{\sqrt{2}}$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{u^2}{2}}$.

Thus, $\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{2}}{u}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1 - \frac{u^2}{2}}}{\frac{u}{\sqrt{2}}} = \frac{\sqrt{2-u^2}}{u}$.

Substitute back:

\[ -\ln \left| \frac{\sqrt{2}}{u} + \frac{\sqrt{2-u^2}}{u} \right| + C = -\ln \left| \frac{\sqrt{2} + \sqrt{2-u^2}}{u} \right| + C \]