Question

Evaluate each integral. $$\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta$$

Short answer

Question: Find the value of the integral $\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta$. Answer: The value of the integral $\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta$ is $\frac{1}{6}(\ln|3+\sin\theta|-\ln|3-\sin\theta|) + C$.

Step-by-step solution

Substitution

Let \(u=\sin\theta\). Then, \(du=\cos\theta\,d\theta\). Substitute these expressions into the integral: $$\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta = \int \frac{1}{9-u^2} du$$

Decompose the integrand into partial fractions

Factor the denominator \(9-u^2 =(3-u)(3+u)\), and express the fraction as a sum of simpler fractions: $$\frac{1}{9-u^2} = \frac{A}{3-u} + \frac{B}{3+u}$$ To solve for \(A\) and \(B\), we multiply both sides by \(9-u^2\), and get: $$1 = A(3+u) + B(3-u)$$ Solve the equation by substituting convenient values for u. For instance, let \(u=3\) and \(u=-3\): $$u = 3: \quad 1 = 6A \implies A = \frac{1}{6}$$ $$u = -3: \quad 1 = -6B \implies B = -\frac{1}{6}$$ Now our integral becomes: $$\int \frac{1}{9-u^2} du = \int\left(\frac{1}{6}\cdot\frac{1}{3-u} -\frac{1}{6}\cdot\frac{1}{3+u}\right) du$$

Evaluate the integral

Rewrite the integral as the sum of two integrals and compute them separately: $$\int \left(\frac{1}{6}\cdot\frac{1}{3-u} -\frac{1}{6}\cdot\frac{1}{3+u}\right) du = \frac{1}{6}\int\frac{1}{3-u}du - \frac{1}{6}\int\frac{1}{3+u}du$$ Use the formula for the integral of the reciprocal function: $$\int \frac{1}{a-u}du = -\ln |a-u| + C_1$$ and $$\int \frac{1}{a+u}du = \ln |a+u| + C_2$$ Hence, the two integrals in our expression become: $$-\frac{1}{6}\ln |3-u| + C_1$$ and $$\frac{1}{6}\ln |3+u| + C_2$$

Combine the results

Add the two integrals to obtain the final result: $$-\frac{1}{6}\ln |3-u| + \frac{1}{6}\ln |3+u| + C = \frac{1}{6}(\ln|3+u|-\ln|3-u|) + C$$

Reverse substitution

Now, replace \(u\) with \(\sin\theta\) to obtain the solution in terms of \(\theta\): $$\frac{1}{6}(\ln|3+\sin\theta|-\ln|3-\sin\theta|) + C$$ So, the final result is: $$\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta = \frac{1}{6}(\ln|3+\sin\theta|-\ln|3-\sin\theta|) + C$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration of expressions involving square roots. It's particularly useful when dealing with integrals that contain expressions like \(a^2 - x^2\), \(a^2 + x^2\), and \(x^2 - a^2\). By using trigonometric identities, you can transform these expressions into simpler forms.

In the original exercise, we set \(u = \sin \theta\), which simplifies the integral. This substitution takes advantage of the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\). Since \(\sin \theta\) already appeared in the denominator, using it as a substitute aligns closely with the desired form \(9 - \sin^2 \theta\).

This approach transforms the integral into a rational function in terms of \(u\), making it easier for further manipulation using partial fraction decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic method used to write a complex rational expression as a sum of simpler fractions. This makes integration much easier. When dealing with a rational function \(\frac{1}{9-u^2}\), recognizing it as a difference of squares is crucial.

• The expression \(9-u^2\) factors into \((3-u)(3+u)\).
  
• By expressing \(\frac{1}{9-u^2}\) as \(\frac{A}{3-u} + \frac{B}{3+u}\), you break the problem into easier parts to integrate.

Finding the constants \(A\) and \(B\) involves algebraic techniques. Substituting values for \(u\) or comparing coefficients will yield \(A = \frac{1}{6}\) and \(B = -\frac{1}{6}\). These constants represent the weights or contribution of each partial fraction in the original function.

This decomposition simplifies the integration process by turning a difficult expression into two separate, more straightforward integrals.

Integral Evaluation

Integral evaluation involves computing the antiderivative of a function. Once you've decomposed into partial fractions, integrating becomes more manageable as each part can be handled using standard integration techniques.

With \(\frac{1}{6}\int \frac{1}{3-u} du\) and \(-\frac{1}{6}\int \frac{1}{3+u} du\), apply the formula for the integral of a reciprocal function: \(\int \frac{1}{a-u} du = -\ln|a-u|\).

• The antiderivatives become \(-\frac{1}{6}\ln|3-u|\) and \(\frac{1}{6}\ln|3+u|\).

After evaluating these, combine them, yielding the result \(\frac{1}{6}(\ln|3+u| - \ln|3-u|) + C\). This represents the primitive function of the original integrand.

Reverse substitution is the final step, replacing \(u\) with \(\sin \theta\) to express the integral in terms of the original variable. Thus, the solution is: \(\frac{1}{6}(\ln|3+\sin \theta| - \ln|3-\sin \theta|) + C\).