What is the integral of sin(mx)sin(nx) ?

The integral of sin(mx)sin(nx) is 1/2 (-1/(m+n)sin(mx+nx)+1/(m-n)sin(mx-nx))+C

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integral of sin(mx)sin(nx)

\int sin (m x) sin (n x) d x

\frac{1}{2} (- \frac{1}{m + n} sin (m x + n x) + \frac{1}{m - n} sin (m x - n x)) + C

Solution steps

\int sin (m x) sin (n x) d x

Rewrite using trig identities

= \int \frac{- cos ( m x + n x ) + cos ( m x - n 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 (m x + n x) + cos (m x - n 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 (m x + n x) d x + \int cos (m x - n x) d x)

\int cos (m x + n x) d x = \frac{1}{m + n} sin (m x + n x)

\int cos (m x - n x) d x = \frac{1}{m - n} sin (m x - n x)

= \frac{1}{2} (- \frac{1}{m + n} sin (m x + n x) + \frac{1}{m - n} sin (m x - n x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{m + n} sin (m x + n x) + \frac{1}{m - n} sin (m x - n x)) + C

y = \frac{8 x ^{2} + 2 x + 6}{x}

\frac{d}{d x} (1000 x e^{- 0.5 x})

\frac{\partial}{\partial x} (z^{3} ln (x y))

f (x) = - 2 x^{2} - 5 x + 2

\int_{0}^{2} 4^{- θ} d θ

What is the integral of sin(mx)sin(nx) ?
The integral of sin(mx)sin(nx) is 1/2 (-1/(m+n)sin(mx+nx)+1/(m-n)sin(mx-nx))+C