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

解

\int_{0}^{2 π} (1 + cos (θ))^{2} d θ

3 π

十進法表記

9.42477 \dots

解答ステップ

\int_{0}^{2 π} (1 + cos (θ))^{2} d θ

拡張

(1 + cos (θ))^{2} : 1 + 2 cos (θ) + cos^{2} (θ)

= \int_{0}^{2 π} 1 + 2 cos (θ) + 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

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

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} 2 cos (θ) d θ = 0

\int_{0}^{2 π} cos^{2} (θ) d θ = π

= 2 π + 0 + π

簡素化

= 3 π