What is the integral of sin(a*x)*sin(b*x) ?

The integral of sin(a*x)*sin(b*x) is 1/2 (-1/(a+b)sin(ax+bx)+1/(a-b)sin(ax-bx))+C

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integral of sin(ax)sin(b*x)

\int sin (a \cdot x) \cdot sin (b \cdot x) d x

\frac{1}{2} (- \frac{1}{a + b} sin (a x + b x) + \frac{1}{a - b} sin (a x - b x)) + C

Solution steps

\int sin (a x) sin (b x) d x

Rewrite using trig identities

= \int \frac{- cos ( a x + b x ) + cos ( a x - b 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 (a x + b x) + cos (a x - b 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 (a x + b x) d x + \int cos (a x - b x) d x)

\int cos (a x + b x) d x = \frac{1}{a + b} sin (a x + b x)

\int cos (a x - b x) d x = \frac{1}{a - b} sin (a x - b x)

= \frac{1}{2} (- \frac{1}{a + b} sin (a x + b x) + \frac{1}{a - b} sin (a x - b x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{a + b} sin (a x + b x) + \frac{1}{a - b} sin (a x - b x)) + C

What is the integral of sin(a*x)*sin(b*x) ?
The integral of sin(a*x)*sin(b*x) is 1/2 (-1/(a+b)sin(ax+bx)+1/(a-b)sin(ax-bx))+C