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

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

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integral of 4cos^3(4x)*sin(4x)

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int cos^{3} (4 x) sin (4 x) d x

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

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

Substitute back

u = cos^{3} (4 x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{( x ^{2} + 1 ) d y}{d x} + 9 x (y - 1) = 0

\frac{d}{d x} (tan^{2} (x) - 1)

\int \frac{6}{x ^{2} x ^{2} - 9} d x

\frac{d}{d x} (5^{x - 2})

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 4

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