What is the integral of cos^3(4θ)sin(4θ) ?

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

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integral of cos^3(4θ)sin(4θ)

\int cos^{3} (4 θ) sin (4 θ) d θ

- \frac{1}{16} cos^{4} (4 θ) + C

Solution steps

\int cos^{3} (4 θ) sin (4 θ) d θ

Apply u-substitution

: \int - \frac{3 u}{12} d u

= \int - \frac{3 u}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int 3 u d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= - \frac{1}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = cos^{3} (4 θ)

= - \frac{1}{12} \cdot \frac{3}{4} (cos^{3} (4 θ))^{\frac{4}{3}}

Simplify

- \frac{1}{12} \cdot \frac{3}{4} (cos^{3} (4 θ))^{\frac{4}{3}} : - \frac{1}{16} cos^{4} (4 θ)

= - \frac{1}{16} cos^{4} (4 θ)

Add a constant to the solution

= - \frac{1}{16} cos^{4} (4 θ) + C

\frac{d y}{d x} = - x e^{- x + y}

y^{^{'}} = y (2 + y) t

\frac{d}{d x} (sin^{x} (x))

\int \frac{1}{x x ^{2} + 1} d x

\frac{d}{d x} (ln (2 x - 4))

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