What is the integral of 9sin^2(θ) ?

The integral of 9sin^2(θ) is 9/2 (θ-1/2 sin(2θ))+C

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integral of 9sin^2(θ)

\int 9 sin^{2} (θ) d θ

\frac{9}{2} (θ - \frac{1}{2} sin (2 θ)) + C

Solution steps

\int 9 sin^{2} (θ) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int sin^{2} (θ) d θ

Rewrite using trig identities

= 9 \cdot \int \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \frac{1}{2} \cdot \int 1 - cos (2 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

= 9 \cdot \frac{1}{2} (\int 1 d θ - \int cos (2 θ) d θ)

\int 1 d θ = θ

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

= 9 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

9 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : \frac{9}{2} (θ - \frac{1}{2} sin (2 θ))

= \frac{9}{2} (θ - \frac{1}{2} sin (2 θ))

Add a constant to the solution

= \frac{9}{2} (θ - \frac{1}{2} sin (2 θ)) + C

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What is the integral of 9sin^2(θ) ?
The integral of 9sin^2(θ) is 9/2 (θ-1/2 sin(2θ))+C