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

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

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integral of 2sin(2t)cos(2t)

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

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 t) d t

Apply u-substitution

= 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 t

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

Simplify

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

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

Add a constant to the solution

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

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