What is the integral from 0 to pi of 1-cos^2(x) ?

The integral from 0 to pi of 1-cos^2(x) is pi/2

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integral from 0 to pi of 1-cos^2(x)

\int_{0}^{π} 1 - cos^{2} (x) d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{π} 1 - cos^{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}

\frac{d}{d x} (8 e^{π x} + 2 e^{7 x})

\frac{\partial}{\partial x} (x cos (y) - y e^{x})

n = 1 \sum \infty \frac{5}{n}

\int e x^{2} d x

e^{\frac{t}{9}} cos (\frac{t}{9})

What is the integral from 0 to pi of 1-cos^2(x) ?
The integral from 0 to pi of 1-cos^2(x) is pi/2