integral of (8x+3sin(x))^2

解

\int (8 x + 3 sin (x))^{2} d x

\frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

解答ステップ

\int (8 x + 3 sin (x))^{2} d x

拡張

(8 x + 3 sin (x))^{2} : 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x)

= \int 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x) 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 64 x^{2} d x + \int 48 x sin (x) d x + \int 9 sin^{2} (x) d x

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

\int 48 x sin (x) d x = 48 (- x cos (x) + sin (x))

\int 9 sin^{2} (x) d x = \frac{9}{2} (x - \frac{1}{2} sin (2 x))

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x))

定数を解答に追加する

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C