What is the integral from 0 to pi of 16sin^2(x) ?

The integral from 0 to pi of 16sin^2(x) is 8pi

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integral from 0 to pi of 16sin^2(x)

\int_{0}^{π} 16 sin^{2} (x) d x

8 π

Decimal

25.13274 \dots

Solution steps

\int_{0}^{π} 16 sin^{2} (x) d x

Take the constant out:

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

= 16 \cdot \int_{0}^{π} sin^{2} (x) d x

Rewrite using trig identities

= 16 \cdot \int_{0}^{π} \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{2} (\int_{0}^{π} 1 d x - \int_{0}^{π} cos (2 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (2 x) d x = 0

= 16 \cdot \frac{1}{2} (π - 0)

Simplify

16 \cdot \frac{1}{2} (π - 0) : 8 π

= 8 π

What is the integral from 0 to pi of 16sin^2(x) ?
The integral from 0 to pi of 16sin^2(x) is 8pi