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

Take the constant out:

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

Apply the Power Rule

= - \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

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