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

解

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

π

十進法表記

3.14159 \dots

解答ステップ

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

定数を除く:

\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

総和規則を適用する:

\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)

簡素化

\frac{1}{2} (2 π + 0) : π

= π