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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\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}

\int_{- 5}^{\infty} e^{- 0.2 x} d x

\int_{1}^{\infty} x^{- \frac{1}{1}} d x

\int_{0}^{5} (2 x - 5) d x

\int_{1}^{8} \frac{1}{x} - \frac{1}{x ^{2}} d x

\int_{- 2 π}^{0} - 36 d t

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