integral of (1+cos(2θ))/2

解

\int \frac{1 + cos ( 2 θ )}{2} d θ

\frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) + C

解答ステップ

\int \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 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 1 d θ + \int cos (2 θ) d θ)

\int 1 d θ = θ

\int cos (2 θ) d θ = \frac{1}{2} sin (2 θ)

= \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))

定数を解答に追加する

= \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) + C