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

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

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

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

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

Solution steps

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

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 t) sin (4 t) d t

Simplify

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

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 t)

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

Simplify

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

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

Add a constant to the solution

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

\int sin^{4} (x) cos^{7} (x) d x

\int 4 x - 11 d x

y^{^{'}} = e^{4 c o s (x)}

2

x \to 6 - lim (36 - x^{2})

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