What is the integral of sin(2θ)sin(6θ) ?

The integral of sin(2θ)sin(6θ) is 1/2 (1/4 sin(4θ)-1/8 sin(8θ))+C

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integral of sin(2θ)sin(6θ)

\int sin (2 θ) sin (6 θ) d θ

\frac{1}{2} (\frac{1}{4} sin (4 θ) - \frac{1}{8} sin (8 θ)) + C

Solution steps

\int sin (2 θ) sin (6 θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} (- cos (8 θ) + cos (- 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 - cos (8 θ) + cos (- 4 θ) d θ

Simplify

- cos (8 θ) + cos (- 4 θ) : cos (4 θ) - cos (8 θ)

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

= \frac{1}{2} (\int cos (4 θ) d θ - \int cos (8 θ) d θ)

\int cos (4 θ) d θ = \frac{1}{4} sin (4 θ)

\int cos (8 θ) d θ = \frac{1}{8} sin (8 θ)

= \frac{1}{2} (\frac{1}{4} sin (4 θ) - \frac{1}{8} sin (8 θ))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} sin (4 θ) - \frac{1}{8} sin (8 θ)) + C

2 sin (3 t) cos (3 t)

\int (z + 3)^{4} d z

(3 x - 2)^{3} (8 x^{2} + 7)^{3}

- \frac{1}{2} x^{3} + 2 x^{2} + \frac{15}{2} x - 9, [1, 6]

\int_{- 1}^{1} 4 - 4 x^{2} d x

What is the integral of sin(2θ)sin(6θ) ?
The integral of sin(2θ)sin(6θ) is 1/2 (1/4 sin(4θ)-1/8 sin(8θ))+C