What is the integral of sin^2(3θ) ?

The integral of sin^2(3θ) is 1/2 (θ-1/6 sin(6θ))+C

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integral of sin^2(3θ)

\int sin^{2} (3 θ) d θ

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

Solution steps

\int sin^{2} (3 θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (6 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 (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

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

\int 1 d θ = θ

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

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

Add a constant to the solution

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

\int - \frac{3}{x} + 3 e^{x} d x

\int (\frac{sin ( x )}{2 e ^{x}}) d x

y^{^{'}} = 5 y^{2} + x y^{2}

\frac{1}{2} ln (1 + x^{2})

n \to \infty lim (n^{- 2} e^{3 n})

What is the integral of sin^2(3θ) ?
The integral of sin^2(3θ) is 1/2 (θ-1/6 sin(6θ))+C