What is the integral of sin(4t)sin(2t) ?

The integral of sin(4t)sin(2t) is 1/2 (-1/6 sin(6t)+1/2 sin(2t))+C

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integral of sin(4t)sin(2t)

\int sin (4 t) sin (2 t) d t

\frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)) + C

Solution steps

\int sin (4 t) sin (2 t) d t

Rewrite using trig identities

= \int \frac{1}{2} (- cos (6 t) + cos (2 t)) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int - cos (6 t) + cos (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int cos (6 t) d t + \int cos (2 t) d t)

\int cos (6 t) d t = \frac{1}{6} sin (6 t)

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

= \frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)) + C

What is the integral of sin(4t)sin(2t) ?
The integral of sin(4t)sin(2t) is 1/2 (-1/6 sin(6t)+1/2 sin(2t))+C