What is the integral of (1-cos(2x))^3 ?

The integral of (1-cos(2x))^3 is 5/2 x-2sin(2x)+3/8 sin(4x)+1/6 sin^3(2x)+C

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integral of (1-cos(2x))^3

\int (1 - cos (2 x))^{3} d x

\frac{5}{2} x - 2 sin (2 x) + \frac{3}{8} sin (4 x) + \frac{1}{6} sin^{3} (2 x) + C

Solution steps

\int (1 - cos (2 x))^{3} d x

Expand

(1 - cos (2 x))^{3} : 1 - 3 cos (2 x) + 3 cos^{2} (2 x) - cos^{3} (2 x)

= \int 1 - 3 cos (2 x) + 3 cos^{2} (2 x) - cos^{3} (2 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 1 d x - \int 3 cos (2 x) d x + \int 3 cos^{2} (2 x) d x - \int cos^{3} (2 x) d x

\int 1 d x = x

\int 3 cos (2 x) d x = \frac{3}{2} sin (2 x)

\int 3 cos^{2} (2 x) d x = \frac{3}{2} (x + \frac{1}{4} sin (4 x))

\int cos^{3} (2 x) d x = \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3})

= x - \frac{3}{2} sin (2 x) + \frac{3}{2} (x + \frac{1}{4} sin (4 x)) - \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3})

Simplify

x - \frac{3}{2} sin (2 x) + \frac{3}{2} (x + \frac{1}{4} sin (4 x)) - \frac{1}{2} (sin (2 x) - \frac{sin ^{3} ( 2 x )}{3}) : \frac{5}{2} x - 2 sin (2 x) + \frac{3}{8} sin (4 x) + \frac{1}{6} sin^{3} (2 x)

= \frac{5}{2} x - 2 sin (2 x) + \frac{3}{8} sin (4 x) + \frac{1}{6} sin^{3} (2 x)

Add a constant to the solution

= \frac{5}{2} x - 2 sin (2 x) + \frac{3}{8} sin (4 x) + \frac{1}{6} sin^{3} (2 x) + C

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What is the integral of (1-cos(2x))^3 ?
The integral of (1-cos(2x))^3 is 5/2 x-2sin(2x)+3/8 sin(4x)+1/6 sin^3(2x)+C