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

The integral of sin(4t)cos(4t) is -1/8 cos^2(4t)+C

Popular >

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

\int sin (4 t) cos (4 t) d t

- \frac{1}{8} cos^{2} (4 t) + C

Solution steps

\int sin (4 t) cos (4 t) d t

Apply u-substitution

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

= \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} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = cos (4 t)

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

Simplify

- \frac{1}{4} \cdot \frac{cos ^{2} ( 4 t )}{2} : - \frac{1}{8} cos^{2} (4 t)

= - \frac{1}{8} cos^{2} (4 t)

Add a constant to the solution

= - \frac{1}{8} cos^{2} (4 t) + C

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

\frac{d ^{2}}{d x ^{2}} (e^{\frac{1}{x}})

\frac{d y}{d x} = \frac{y}{x ^{2}}, y (1) = 2

arctan (sin (2 x))

f (x) = x, at x = \frac{1}{81}

What is the integral of sin(4t)cos(4t) ?
The integral of sin(4t)cos(4t) is -1/8 cos^2(4t)+C