What is the integral of 2sin(2θ)sin(4θ) ?

The integral of 2sin(2θ)sin(4θ) is 1/2 sin(2θ)-1/6 sin(6θ)+C

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integral of 2sin(2θ)sin(4θ)

\int 2 sin (2 θ) sin (4 θ) d θ

\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ) + C

Solution steps

\int 2 sin (2 θ) sin (4 θ) d θ

Take the constant out:

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

= 2 \cdot \int sin (2 θ) sin (4 θ) d θ

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2} (- cos (6 θ) + cos (- 2 θ)) d θ

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 - cos (6 θ) + cos (- 2 θ) d θ

Simplify

- cos (6 θ) + cos (- 2 θ) : cos (2 θ) - cos (6 θ)

= 2 \cdot \frac{1}{2} \cdot \int cos (2 θ) - cos (6 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{2} (\int cos (2 θ) d θ - \int cos (6 θ) d θ)

\int cos (2 θ) d θ = \frac{1}{2} sin (2 θ)

\int cos (6 θ) d θ = \frac{1}{6} sin (6 θ)

= 2 \cdot \frac{1}{2} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ))

Simplify

2 \cdot \frac{1}{2} (\frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)) : \frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)

= \frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ)

Add a constant to the solution

= \frac{1}{2} sin (2 θ) - \frac{1}{6} sin (6 θ) + C

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What is the integral of 2sin(2θ)sin(4θ) ?
The integral of 2sin(2θ)sin(4θ) is 1/2 sin(2θ)-1/6 sin(6θ)+C