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

The integral from 0 to 2pi of (1+cos(2θ))/2 is pi

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integral from 0 to 2pi of (1+cos(2θ))/2

\int_{0}^{2 π} \frac{1 + cos ( 2 θ )}{2} d θ

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{2 π} \frac{1 + cos ( 2 θ )}{2} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 π} 1 + cos (2 θ) d θ

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}^{2 π} 1 d θ + \int_{0}^{2 π} cos (2 θ) d θ)

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} cos (2 θ) d θ = 0

= \frac{1}{2} (2 π + 0)

Simplify

\frac{1}{2} (2 π + 0) : π

= π

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