What is the integral of sin(3x)sin(kx) ?

The integral of sin(3x)sin(kx) is 1/2 (-1/(3+k)sin(3x+kx)+1/(3-k)sin(3x-kx))+C

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integral of sin(3x)sin(kx)

\int sin (3 x) sin (k x) d x

\frac{1}{2} (- \frac{1}{3 + k} sin (3 x + k x) + \frac{1}{3 - k} sin (3 x - k x)) + C

Solution steps

\int sin (3 x) sin (k x) d x

Rewrite using trig identities

= \int \frac{- cos ( 3 x + k x ) + cos ( 3 x - k x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int - cos (3 x + k x) + cos (3 x - k x) d x

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 (3 x + k x) d x + \int cos (3 x - k x) d x)

\int cos (3 x + k x) d x = \frac{1}{3 + k} sin (3 x + k x)

\int cos (3 x - k x) d x = \frac{1}{3 - k} sin (3 x - k x)

= \frac{1}{2} (- \frac{1}{3 + k} sin (3 x + k x) + \frac{1}{3 - k} sin (3 x - k x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{3 + k} sin (3 x + k x) + \frac{1}{3 - k} sin (3 x - k x)) + C

What is the integral of sin(3x)sin(kx) ?
The integral of sin(3x)sin(kx) is 1/2 (-1/(3+k)sin(3x+kx)+1/(3-k)sin(3x-kx))+C