What is the integral of sin^2(4θ) ?

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

integral of sin^2(4θ)

\int sin^{2} (4 θ) d θ

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

Solution steps

\int sin^{2} (4 θ) d θ

Rewrite using trig identities

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

\int 1 d θ = θ

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

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

Add a constant to the solution

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

l a pl a ce t r an s f or m f (t) = 5 cos (8 t) + 3 e^{- 4} sin (3 t)

\frac{d}{d x} ((x^{3} + 1)^{4})

\int sec (x) - cos (x) d x

d er i v a t i v e \frac{1}{2} ln (x)

\int x ln (11 x) d x

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