What is the integral from 0 to pi of rsin^2(θ) ?

The integral from 0 to pi of rsin^2(θ) is pi/2 r

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integral from 0 to pi of rsin^2(θ)

\int_{0}^{π} r sin^{2} (θ) d θ

\frac{π}{2} r

Solution steps

\int_{0}^{π} r sin^{2} (θ) d θ

Take the constant out:

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

= r \cdot \int_{0}^{π} sin^{2} (θ) d θ

Rewrite using trig identities

= r \cdot \int_{0}^{π} \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

= r \frac{1}{2} \cdot \int_{0}^{π} 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

= r \frac{1}{2} (\int_{0}^{π} 1 d θ - \int_{0}^{π} cos (2 θ) d θ)

\int_{0}^{π} 1 d θ = π

\int_{0}^{π} cos (2 θ) d θ = 0

= r \frac{1}{2} (π - 0)

Simplify

r \frac{1}{2} (π - 0) : \frac{π}{2} r

= \frac{π}{2} r

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What is the integral from 0 to pi of rsin^2(θ) ?
The integral from 0 to pi of rsin^2(θ) is pi/2 r