Question
Using Product-to-Sum Identities In Exercises $41-46$ find the indefinite integral.$$\int \sin \theta \sin 3 \theta d \theta$$
Step 1
The identity for $\sin A \sin B$ is $\frac{1}{2}[\cos(A-B)-\cos(A+B)]$. Here, $A=\theta$ and $B=3\theta$. So, we get $$ \int \sin \theta \sin 3 \theta d \theta = \frac{1}{2} \int [\cos(\theta-3\theta)-\cos(\theta+3\theta)] d\theta $$ Show more…
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