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

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

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

\int (sin (2 x) cos (2 x)) d x

- \frac{1}{8} cos (4 x) + C

Solution steps

\int sin (2 x) cos (2 x) d x

Rewrite using trig identities

= \int \frac{1}{2} sin (4 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (4 x) d x

Apply u-substitution

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

= \frac{1}{2} \cdot \int 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}{2} \cdot \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

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

Substitute back

u = 4 x

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

Simplify

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

= - \frac{1}{8} cos (4 x)

Add a constant to the solution

= - \frac{1}{8} cos (4 x) + C

\frac{d}{d x} (e^{x^{3} y})

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

\int_{1}^{3} \frac{( 3 x ^{2} - 7 )}{x ^{3}} d x

f (x) = x^{3} e^{4 x}

y^{^{'}} + x (y^{2} + y) = 0, y (2) = 1

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