Problem 35 - Essential Calculus

Question

Find or evaluate the integral. $$ \int \frac{\sin \theta}{3-2 \cos \theta} d \theta $$

Step-by-Step Solution

Answer

The integral of \( \int \frac{\sin \theta}{3-2 \cos \theta} d \theta \) is \(-\frac{1}{2} \ln |3 - 2 \cos \theta|\)

Step 1: Change of Variable

A substitution will simplify this equation greatly. Let's choose the substitution \(x = \cos \theta\). This gives us \(dx = -\sin \theta \, d\theta\). Now we can rewrite the integral in terms of \(x\). The new integral will be: \[-\int \frac{dx}{3 - 2x}\]

Step 2: Evaluate the Integral

Now, we can solve the integral. This is a simple rational function, the antiderivative of this function is the natural logarithm. We can write the integral \( -\int \frac{dx}{3-2x} \) as \(-\frac{1}{2}\int \frac{d(3-2x)}{3-2x}\), which is \(-\frac{1}{2} \ln |3-2x|\)

Step 3: Substituting Back

Now, we substitute \(x = \cos \theta\) back into the result to get the final answer: \(-\frac{1}{2} \ln |3 - 2 \cos \theta|\)