Problem 90 - Calculus Early Transcendentals

Question

Evaluate the following integrals. \(\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta\)

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \frac{\cos \theta}{9-\sin^2 \theta} d\theta\). Answer: The evaluated integral is \(\int \frac{\cos \theta}{9-\sin^2 \theta} d\theta = \frac{1}{6} \operatorname{arctanh}(\sin \theta) + C\), where \(C\) is the constant of integration.

Step 1: Choose a suitable substitution

We choose the substitution \(x = 3\sin \theta\). This will help simplify the denominator of the integrand. Now, we need to find the differential \(dx\).

Step 2: Finding the differential

To find the differential, we differentiate both sides of the substitution equation with respect to \(\theta\): $$ \frac{dx}{d\theta} = 3\cos \theta $$ Therefore, \(d\theta=\frac{1}{3\cos \theta} dx\)

Step 3: Perform the substitution

Substitute \(x\) and \(d\theta\) in the integral, and replace \(\cos \theta\) with the expression from the substitution equation: $$ \int \frac{\cos \theta}{9-\sin^2 \theta} d\theta = \int \frac{\cos \theta}{9-x^2} \cdot \frac{1}{3\cos \theta} dx $$ Simplify the integral: $$ = \frac{1}{3} \int \frac{dx}{9-x^2} $$

Step 4: Integrate the simplified integral

The simplified integral is now a standard form which can be integrated using the inverse hyperbolic functions. The integral becomes: $$ \frac{1}{3} \int \frac{dx}{9-x^2} = \frac{1}{6} \operatorname{arctanh}\left(\frac{x}{3}\right) + C $$ where C is the constant of integration.

Step 5: Express the result in terms of the original variable

Finally, substitute back the original variable \(\theta\) using the substitution equation \(x=3\sin \theta\): $$ \frac{1}{6} \operatorname{arctanh}\left(\frac{3\sin \theta}{3}\right) + C = \frac{1}{6} \operatorname{arctanh}(\sin \theta) + C $$ So, the evaluated integral is: $$ \int \frac{\cos \theta}{9-\sin^2 \theta} d\theta = \frac{1}{6} \operatorname{arctanh}(\sin \theta) + C $$

Key Concepts

Substitution Method

Understanding the substitution method is important for simplifying complex integrals. The key idea is to transform an integral into a more manageable form.

Here's how it works:

• Identify a substitution that could simplify the expression.
• Replace the variable with the new substitution.
• Adjust the differential accordingly.

In our example, we used the substitution \(x = 3\sin \theta\). This choice makes the expression in the denominator, \(9 - \sin^2 \theta\), easier to handle, turning it into \(9 - x^2\).

This process reduces complex functions into integrals that are simpler to evaluate, often leading to standard forms with known solutions.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are related to hyperbolic functions just as inverse trigonometric functions are related to trigonometric functions.

These functions often appear in integration when simplifying integrals involving quadratic expressions.

In this context:

• \(\operatorname{arctanh}\) is the inverse hyperbolic tangent function.
• They frequently arise when integrating functions of the form \(\int \frac{dx}{a^2 - x^2}\).

In the given exercise, we reached the integral \(\frac{1}{3} \int \frac{dx}{9-x^2}\), which can be solved using the inverse hyperbolic tangent, \(\operatorname{arctanh}\), resulting in a form that matches known identities for integrating these expressions.

Trigonometric Integration

Trigonometric integration involves integrals that include trigonometric functions like \(\sin \theta\) and \(\cos \theta\).

Such integrals sometimes look complicated but often have symmetries or substitution opportunities to make them manageable.

In solving \(\int \frac{\cos \theta}{9-\sin^2 \theta} d \theta\):

• Recognize the presence of trigonometric identities.
• Look for substitutions that align with these identities, like transforming \(9 - \sin^2 \theta\) using \(x = 3\sin \theta\).

These approaches not only simplify the integrals but also make them accessible to standard solution techniques, like inverse hyperbolic functions, connecting trigonometric functions to broader mathematical concepts.