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

The integral from 0 to pi/2 of-sin^2(x) is -pi/4

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

\int_{0}^{\frac{π}{2}} - sin^{2} (x) d x

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

\int_{0}^{\frac{π}{2}} - 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}^{\frac{π}{2}} sin^{2} (x) d x

Rewrite using trig identities

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

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 x) d x = 0

= - \frac{1}{2} (\frac{π}{2} - 0)

Simplify

- \frac{1}{2} (\frac{π}{2} - 0) : - \frac{π}{4}

= - \frac{π}{4}

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