What is the integral of sin^2(4θ) ?

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

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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

\int \frac{7 x ^{2} + 3 x + 7}{( x ^{2} + 1 ) ^{2}} d x

f (t) = 5 cos (8 t) + 3 e^{- 4} sin (3 t)

x \to 0 lim (ln (x) - x)

\frac{\partial}{\partial x} (y + z)

\frac{d}{d x} (cot (x) + arctan (x + 1))

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