integral of (sin(x)+1)^2

解

\int (sin (x) + 1)^{2} d x

\frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C

解答ステップ

\int (sin (x) + 1)^{2} d x

拡張

(sin (x) + 1)^{2} : sin^{2} (x) + 2 sin (x) + 1

= \int sin^{2} (x) + 2 sin (x) + 1 d x

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 sin^{2} (x) d x + \int 2 sin (x) d x + \int 1 d x

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

\int 2 sin (x) d x = - 2 cos (x)

\int 1 d x = x

= \frac{1}{2} (x - \frac{1}{2} sin (2 x)) - 2 cos (x) + x

簡素化

\frac{1}{2} (x - \frac{1}{2} sin (2 x)) - 2 cos (x) + x : \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

定数を解答に追加する

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C