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

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

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

\int_{π}^{0} - sin^{2} (x) d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{π}^{0} - sin^{2} (x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{π} - sin^{2} (x) d x

Take the constant out:

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

= - (- \int_{0}^{π} sin^{2} (x) d x)

Rewrite using trig identities

= - (- \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

= - (- \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

= - (- \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

= - (- \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 pi to 0 of-sin^2(x) ?
The integral from pi to 0 of-sin^2(x) is pi/2