integral of (sin^3(x)-3sin^2(x))(sin^4(x)-4sin^3(x))6cos(x)

解

\int (sin^{3} (x) - 3 sin^{2} (x)) (sin^{4} (x) - 4 sin^{3} (x)) 6 cos (x) d x

6 (\frac{sin ^{8} ( x )}{8} - sin^{7} (x) + 2 sin^{6} (x)) + C

解答ステップ

\int (sin^{3} (x) - 3 sin^{2} (x)) (sin^{4} (x) - 4 sin^{3} (x)) \cdot 6 cos (x) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int (sin^{3} (x) - 3 sin^{2} (x)) (sin^{4} (x) - 4 sin^{3} (x)) cos (x) d x

拡張

(sin^{3} (x) - 3 sin^{2} (x)) (sin^{4} (x) - 4 sin^{3} (x)) cos (x) : sin^{7} (x) cos (x) - 7 sin^{6} (x) cos (x) + 12 sin^{5} (x) cos (x)

= 6 \cdot \int sin^{7} (x) cos (x) - 7 sin^{6} (x) cos (x) + 12 sin^{5} (x) cos (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

= 6 (\int sin^{7} (x) cos (x) d x - \int 7 sin^{6} (x) cos (x) d x + \int 12 sin^{5} (x) cos (x) d x)

\int sin^{7} (x) cos (x) d x = \frac{sin ^{8} ( x )}{8}

\int 7 sin^{6} (x) cos (x) d x = sin^{7} (x)

\int 12 sin^{5} (x) cos (x) d x = 2 sin^{6} (x)

= 6 (\frac{sin ^{8} ( x )}{8} - sin^{7} (x) + 2 sin^{6} (x))

定数を解答に追加する

= 6 (\frac{sin ^{8} ( x )}{8} - sin^{7} (x) + 2 sin^{6} (x)) + C