Question
Evaluate the integral.$\int \sin (-4 x) \cos (-2 x) d x$
Step 1
The identity we will use is $\sin(a) \cos(b) = \frac{1}{2}[\sin(a+b) + \sin(a-b)]$. Applying this identity to our integral, we get: \[\int \sin(-4x) \cos(-2x) dx = \frac{1}{2} \int [\sin((-4x) + (-2x)) + \sin((-4x) - (-2x))] dx\] Show more…
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