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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int cos^{2} (u) sin (u) \frac{1}{4} d u

= \int cos^{2} (u) sin (u) \frac{1}{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 cos^{2} (u) sin (u) d u

Apply u-substitution

: \int - v^{2} d v

= \frac{1}{4} \cdot \int - v^{2} d v

Take the constant out:

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

= \frac{1}{4} (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{4} (- \frac{v ^{3}}{3})

Substitute back

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

Simplify

\frac{1}{4} (- \frac{cos ^{3} ( 4 x )}{3}) : - \frac{1}{12} cos^{3} (4 x)

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} (x^{2} + y^{2} - 2)

\int \frac{1}{x ^{2} x ^{2} - 4} d x

\int \frac{2}{x ( x ^{4} + 25 ) ^{\frac{1}{2}}} d x

h \to 2 lim (\frac{h ^{3} + 8}{h + 2})

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