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

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

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integral of sin(nx)sin(mx)

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

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

Solution steps

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

Rewrite using trig identities

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

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

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

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

Add a constant to the solution

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

f (x) = \frac{3}{2 3 x + 1}, at x = 8

y^{^{'}} = \frac{y}{x} + \frac{x}{2 ( y + x )}

\int_{2}^{5} \frac{x}{x - 1} d x

\frac{d}{d x} (\frac{1}{4 x ^{2} + 1})

\frac{\partial}{\partial x} (\frac{( 2 x + 3 y )}{4 x + 5 y})

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