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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 2 \cdot \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

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{3}{x ^{4}} d x

f (x) = - \frac{x ^{3} + x - 1}{2}

\frac{d}{d x} (e^{2 x} + 2 + e^{- 2 x})

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

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