What is the integral from 0 to pi of 1/2 (4sin(θ))^2 ?

The integral from 0 to pi of 1/2 (4sin(θ))^2 is 4pi

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integral from 0 to pi of 1/2 (4sin(θ))^2

\int_{0}^{π} \frac{1}{2} (4 sin (θ))^{2} d θ

4 π

Decimal

12.56637 \dots

Solution steps

\int_{0}^{π} \frac{1}{2} (4 sin (θ))^{2} 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_{0}^{π} (4 sin (θ))^{2} d θ

Apply Integration By Parts

= \frac{1}{2} [16 θ sin^{2} (θ) - \int 32 θ sin (θ) cos (θ) d θ]_{0}^{π}

\int 32 θ sin (θ) cos (θ) d θ = 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ))

= \frac{1}{2} [16 θ sin^{2} (θ) - 16 (- \frac{1}{2} θ cos (2 θ) + \frac{1}{4} sin (2 θ))]_{0}^{π}

Compute the boundaries:

8 π

= \frac{1}{2} \cdot 8 π

Simplify

= 4 π

What is the integral from 0 to pi of 1/2 (4sin(θ))^2 ?
The integral from 0 to pi of 1/2 (4sin(θ))^2 is 4pi