What is the integral of cos(4x)(1-sin(4x))^2 ?

The integral of cos(4x)(1-sin(4x))^2 is 1/4 sin(4x)+1/4 cos^2(4x)+1/12 sin^3(4x)+C

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integral of cos(4x)(1-sin(4x))^2

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

\frac{1}{4} sin (4 x) + \frac{1}{4} cos^{2} (4 x) + \frac{1}{12} sin^{3} (4 x) + C

Solution steps

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

Expand

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

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

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

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

\int sin^{2} (4 x) cos (4 x) d x = \frac{1}{12} sin^{3} (4 x)

= \frac{1}{4} sin (4 x) - (- \frac{1}{4} cos^{2} (4 x)) + \frac{1}{12} sin^{3} (4 x)

Simplify

= \frac{1}{4} sin (4 x) + \frac{1}{4} cos^{2} (4 x) + \frac{1}{12} sin^{3} (4 x)

Add a constant to the solution

= \frac{1}{4} sin (4 x) + \frac{1}{4} cos^{2} (4 x) + \frac{1}{12} sin^{3} (4 x) + C

\int_{0}^{1} \frac{4}{1 + x ^{2}} d x

\frac{d}{d x} (e^{x y} y)

\int (x + 3) 7 - x d x

x \to 0 lim (x^{4} sin (\frac{3}{x}))

\frac{d}{d x} ((2 x^{3} + 5)^{4})

What is the integral of cos(4x)(1-sin(4x))^2 ?
The integral of cos(4x)(1-sin(4x))^2 is 1/4 sin(4x)+1/4 cos^2(4x)+1/12 sin^3(4x)+C