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人気のある微分積分 >

integral of (sin(2x)-sin(3x))^2

解

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

解

x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{1}{2} sin^{2} (u) - sin (\frac{3 u}{2}) sin (u) + \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u

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 \frac{1}{2} sin^{2} (u) d u - \int sin (\frac{3 u}{2}) sin (u) d u + \int \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u

\int \frac{1}{2} sin^{2} (u) d u = \frac{1}{4} (u - \frac{1}{2} sin (2 u))

\int sin (\frac{3 u}{2}) sin (u) d u = - \frac{1}{5} sin (\frac{5 u}{2}) + sin (\frac{u}{2})

\int \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u = \frac{1}{4} (u - \frac{1}{3} sin (3 u))

= \frac{1}{4} (u - \frac{1}{2} sin (2 u)) - (- \frac{1}{5} sin (\frac{5 u}{2}) + sin (\frac{u}{2})) + \frac{1}{4} (u - \frac{1}{3} sin (3 u))

代用を戻す

u = 2 x

= \frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{1}{5} sin (\frac{5 \cdot 2 x}{2}) + sin (\frac{2 x}{2})) + \frac{1}{4} (2 x - \frac{1}{3} sin (3 \cdot 2 x))

簡素化

\frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{1}{5} sin (\frac{5 \cdot 2 x}{2}) + sin (\frac{2 x}{2})) + \frac{1}{4} (2 x - \frac{1}{3} sin (3 \cdot 2 x)) : x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x)

= x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x)

定数を解答に追加する

= x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x) + C

グラフ

人気の例

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