# How do you find the integral of #Sin^3(mx) dx#?

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To find the integral of (\sin^3(mx) , dx):

- Use the trigonometric identity (\sin^2(x) = 1 - \cos^2(x)) to rewrite (\sin^3(mx)) in terms of (\sin(mx)) and (\cos(mx)).
- Apply the power reduction formula for sine cubed: [\sin^3(x) = \frac{3}{4}\sin(x) - \frac{1}{4}\sin(3x)]
- Replace (x) with (mx) to account for the coefficient (m) in the given integral.
- Integrate each term separately.

After applying the power reduction formula, the integral becomes:

[ \int \sin^3(mx) , dx = \frac{3}{4m} \int \sin(mx) , dx - \frac{1}{4m} \int \sin(3mx) , dx ]

Now, integrate each term separately:

[ \frac{3}{4m} \int \sin(mx) , dx = -\frac{3}{4m} \frac{\cos(mx)}{m} + C_1 ]

[ \frac{1}{4m} \int \sin(3mx) , dx = -\frac{1}{12m} \frac{\cos(3mx)}{m} + C_2 ]

Where (C_1) and (C_2) are constants of integration.

So, the integral of (\sin^3(mx) , dx) is given by:

[ \frac{-3}{4m^2} \cos(mx) - \frac{1}{12m^2} \cos(3mx) + C ]

Where (C) is the constant of integration.

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