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

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

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integral of sin(2θ)cos(2θ)

\int sin (2 θ) cos (2 θ) d θ

- \frac{1}{8} cos (4 θ) + C

Solution steps

\int sin (2 θ) cos (2 θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} sin (4 θ) d θ

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 θ) d θ

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 θ

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

Simplify

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

= - \frac{1}{8} cos (4 θ)

Add a constant to the solution

= - \frac{1}{8} cos (4 θ) + C

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