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

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

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

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

- \frac{1}{16} (1 - sin (4 x))^{4} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u ^{3}}{4} d u

= \int - \frac{u ^{3}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{1}{4} \cdot \frac{u ^{4}}{4}

Substitute back

u = 1 - sin (4 x)

= - \frac{1}{4} \cdot \frac{( 1 - sin ( 4 x ) ) ^{4}}{4}

Simplify

- \frac{1}{4} \cdot \frac{( 1 - sin ( 4 x ) ) ^{4}}{4} : - \frac{1}{16} (1 - sin (4 x))^{4}

= - \frac{1}{16} (1 - sin (4 x))^{4}

Add a constant to the solution

= - \frac{1}{16} (1 - sin (4 x))^{4} + C

x \to - 3 lim (\frac{x ^{2} - 8}{3 - x})

\frac{d}{d x} (e^{x^{2}} - x)

y^{^{'}} = \frac{( x ^{2} + y ^{2} )}{x y}

\frac{d}{d x} (14 x e^{- x})

x^{2} f (x^{2} + 2 x)

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