What is the integral of (-cos(4x)sin(4x))/4 ?

The integral of (-cos(4x)sin(4x))/4 is 1/32 cos^2(4x)+C

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integral of (-cos(4x)sin(4x))/4

\int \frac{- cos ( 4 x ) sin ( 4 x )}{4} d x

\frac{1}{32} cos^{2} (4 x) + C

Solution steps

\int \frac{- cos ( 4 x ) sin ( 4 x )}{4} d x

Take the constant out:

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

= \frac{- 1}{4} \cdot \int cos (4 x) sin (4 x) d x

Simplify

= (- \frac{1}{4}) \cdot \int cos (4 x) sin (4 x) d x

Apply u-substitution

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

= (- \frac{1}{4}) \cdot \int - \frac{u}{4} d u

Take the constant out:

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

= (- \frac{1}{4}) (- \frac{1}{4} \cdot \int u d u)

Apply the Power Rule

: \frac{u ^{2}}{2}

= (- \frac{1}{4}) (- \frac{1}{4} \cdot \frac{u ^{2}}{2})

Substitute back

u = cos (4 x)

= (- \frac{1}{4}) (- \frac{1}{4} \cdot \frac{cos ^{2} ( 4 x )}{2})

Simplify

(- \frac{1}{4}) (- \frac{1}{4} \cdot \frac{cos ^{2} ( 4 x )}{2}) : \frac{1}{32} cos^{2} (4 x)

= \frac{1}{32} cos^{2} (4 x)

Add a constant to the solution

= \frac{1}{32} cos^{2} (4 x) + C

\int_{0}^{6} 2 x + 1 d x

\frac{d}{d t} (- t)

x^{2} - 2 x y + x \frac{d y}{d x} = 0

\frac{2}{s ^{2} ( s ^{2} + 5 )}

\frac{d ^{2}}{d x ^{2}} (16 x^{3} - 34 x^{2})

What is the integral of (-cos(4x)sin(4x))/4 ?
The integral of (-cos(4x)sin(4x))/4 is 1/32 cos^2(4x)+C