What is the integral of sin(4x)(1-cos(4x))^3 ?

The integral of sin(4x)(1-cos(4x))^3 is -1/4 cos(4x)+3/8 cos^2(4x)-1/4 cos^3(4x)+1/16 cos^4(4x)+C

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integral of sin(4x)(1-cos(4x))^3

\int sin (4 x) (1 - cos (4 x))^{3} d x

- \frac{1}{4} cos (4 x) + \frac{3}{8} cos^{2} (4 x) - \frac{1}{4} cos^{3} (4 x) + \frac{1}{16} cos^{4} (4 x) + C

Solution steps

\int sin (4 x) (1 - cos (4 x))^{3} d x

Expand

sin (4 x) (1 - cos (4 x))^{3} : sin (4 x) - 3 sin (4 x) cos (4 x) + 3 cos^{2} (4 x) sin (4 x) - cos^{3} (4 x) sin (4 x)

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

\int sin (4 x) d x = - \frac{1}{4} cos (4 x)

\int 3 sin (4 x) cos (4 x) d x = - \frac{3}{8} cos^{2} (4 x)

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

\int cos^{3} (4 x) sin (4 x) d x = - \frac{1}{16} cos^{4} (4 x)

= - \frac{1}{4} cos (4 x) - (- \frac{3}{8} cos^{2} (4 x)) - \frac{1}{4} cos^{3} (4 x) - (- \frac{1}{16} cos^{4} (4 x))

Simplify

= - \frac{1}{4} cos (4 x) + \frac{3}{8} cos^{2} (4 x) - \frac{1}{4} cos^{3} (4 x) + \frac{1}{16} cos^{4} (4 x)

Add a constant to the solution

= - \frac{1}{4} cos (4 x) + \frac{3}{8} cos^{2} (4 x) - \frac{1}{4} cos^{3} (4 x) + \frac{1}{16} cos^{4} (4 x) + C

What is the integral of sin(4x)(1-cos(4x))^3 ?
The integral of sin(4x)(1-cos(4x))^3 is -1/4 cos(4x)+3/8 cos^2(4x)-1/4 cos^3(4x)+1/16 cos^4(4x)+C