What is the integral from 0 to 3 of sin^2(mx) ?

The integral from 0 to 3 of sin^2(mx) is 3-1/2 (3+(sin(6m))/(2m))

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integral from 0 to 3 of sin^2(mx)

\int_{0}^{3} sin^{2} (m x) d x

3 - \frac{1}{2} (3 + \frac{sin ( 6 m )}{2 m})

Solution steps

\int_{0}^{3} sin^{2} (m x) d x

Rewrite using trig identities

= \int_{0}^{3} 1 - cos^{2} (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

= \int_{0}^{3} 1 d x - \int_{0}^{3} cos^{2} (m x) d x

\int_{0}^{3} 1 d x = 3

\int_{0}^{3} cos^{2} (m x) d x = \frac{1}{2} (3 + \frac{sin ( 6 m )}{2 m})

= 3 - \frac{1}{2} (3 + \frac{sin ( 6 m )}{2 m})

\frac{1}{6 x - 12} 2 x + 16

\frac{d}{d x} (x + \frac{1}{x ^{3}})

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

h (x) = 3 e^{x} - 7 x, (0, 3)

\frac{1}{2 s + 5}

What is the integral from 0 to 3 of sin^2(mx) ?
The integral from 0 to 3 of sin^2(mx) is 3-1/2 (3+(sin(6m))/(2m))