What is the integral from 0 to pi of-sin^2(t) ?

The integral from 0 to pi of-sin^2(t) is -pi/2

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integral from 0 to pi of-sin^2(t)

\int_{0}^{π} - sin^{2} (t) d t

- \frac{π}{2}

Decimal

- 1.57079 \dots

Solution steps

\int_{0}^{π} - sin^{2} (t) d t

Take the constant out:

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

= - \int_{0}^{π} sin^{2} (t) d t

Rewrite using trig identities

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

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} 1 - cos (2 t) d t

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_{0}^{π} 1 d t - \int_{0}^{π} cos (2 t) d t)

\int_{0}^{π} 1 d t = π

\int_{0}^{π} cos (2 t) d t = 0

= - \frac{1}{2} (π - 0)

Simplify

- \frac{1}{2} (π - 0) : - \frac{π}{2}

= - \frac{π}{2}

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What is the integral from 0 to pi of-sin^2(t) ?
The integral from 0 to pi of-sin^2(t) is -pi/2