What is the integral of sin^2(2θ) ?

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

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integral of sin^2(2θ)

\int sin^{2} (2 θ) d θ

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

Solution steps

\int sin^{2} (2 θ) d θ

Rewrite using trig identities

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

\int 1 d θ = θ

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

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

Add a constant to the solution

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

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} + 12 y = 0

\frac{d}{d x} (\frac{4}{3} x^{3} + 2)

\int_{0}^{\infty} \frac{7}{x ^{2} + 1} d x

\int \frac{1}{x ^{8} + x ^{6}} d x

x \to 0 + lim (x (ln (4 x))^{2})

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