Problem 15 - Essential Calculus

Question

Use integration tables to find the integral. $$ \int \frac{\cos \theta}{3+2 \sin \theta+\sin ^{2} \theta} d \theta $$

Step-by-Step Solution

Answer

\[ \frac{1}{\sqrt{5}}\arctan\left(\frac{2\sin\theta+1}{\sqrt{5}}\right) + C \]

Step 1: Identify suitable substitution

First, it's necessary to note that if \(u = \sin \theta\), the expression inside the integral becomes simpler. So, let's let \(u = \sin \theta\), then \(du = \cos \theta d \theta\). This will allow us to rewrite the integral.

Step 2: Apply substitution

Apply the substitution \(u = \sin \theta\), then \(du = \cos \theta d \theta\). So, the original integral can be rewritten as: \[ \int \frac{1}{3+2u+u^{2}} d u \]

Step 3: Use the Integral Tables

The integral is now in a form that can be found using standard integral tables. The integral of \(\frac{1}{3+2u+u^{2}}\) with respect to \(u\) is \(\frac{1}{\sqrt{5}}\arctan\left(\frac{2u+1}{\sqrt{5}}\right) + C\). Here, \(\arctan(x)\) is the inverse tangent function and \(C\) is the constant of integration.

Step 4: Substitute u back in

Now it's time to substitute \(u = \sin \theta\) back into the solution. This yields the final answer of: \[ \frac{1}{\sqrt{5}}\arctan\left(\frac{2\sin\theta+1}{\sqrt{5}}\right) + C \]