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

Short Answer

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

Step by step solution

01

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.
02

Apply substitution

Apply the substitution \(u = \sin \theta\), then \(du = \cos \theta d \theta\). So, the original integral can be rewritten as: \[ \int \frac{1}{3+2u+u^{2}} d u \]
03

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.
04

Substitute u back in

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

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