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) \cdot 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

n = 1 \sum \infty \frac{1}{n} x^{n}

n = 1 \sum \infty (\frac{6}{4})^{n}

\int (1 - x) d x

\frac{\partial}{\partial x} (sin (4 x - 7 y))

\frac{( x + 1 ) d y}{d x} = - y + 10

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