integral of 1+2cos(2θ)+1/2 (1+cos(4θ))

解

\int 1 + 2 cos (2 θ) + \frac{1}{2} (1 + cos (4 θ)) d θ

\frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ) + C

解答ステップ

\int 1 + 2 cos (2 θ) + \frac{1}{2} (1 + cos (4 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 1 d θ + \int 2 cos (2 θ) d θ + \int \frac{1}{2} (1 + cos (4 θ)) d θ

\int 1 d θ = θ

\int 2 cos (2 θ) d θ = sin (2 θ)

\int \frac{1}{2} (1 + cos (4 θ)) d θ = \frac{1}{2} (θ + \frac{1}{4} sin (4 θ))

= θ + sin (2 θ) + \frac{1}{2} (θ + \frac{1}{4} sin (4 θ))

簡素化

θ + sin (2 θ) + \frac{1}{2} (θ + \frac{1}{4} sin (4 θ)) : \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ)

= \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ)

定数を解答に追加する

= \frac{3}{2} θ + \frac{1}{8} sin (4 θ) + sin (2 θ) + C