integral of (cos^7(y)sin^2(y))

解答

\int (cos^{7} (y) sin^{2} (y)) d y

\frac{sin ^{3} ( y )}{3} - \frac{3 sin ^{5} ( y )}{5} + \frac{3 sin ^{7} ( y )}{7} - \frac{sin ^{9} ( y )}{9} + C

求解步骤

\int cos^{7} (y) sin^{2} (y) d y

使用三角恒等式改写

= \int (1 - sin^{2} (y))^{3} cos (y) sin^{2} (y) d y

使用换元积分法

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

乘开

u^{2} (1 - u^{2})^{3} : u^{2} - 3 u^{4} + 3 u^{6} - u^{8}

= \int u^{2} - 3 u^{4} + 3 u^{6} - u^{8} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u - \int 3 u^{4} d u + \int 3 u^{6} d u - \int u^{8} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{3}}{3} - \frac{3 u ^{5}}{5} + \frac{3 u ^{7}}{7} - \frac{u ^{9}}{9}

u = sin (y)

代回

= \frac{sin ^{3} ( y )}{3} - \frac{3 sin ^{5} ( y )}{5} + \frac{3 sin ^{7} ( y )}{7} - \frac{sin ^{9} ( y )}{9}

解答补常数

= \frac{sin ^{3} ( y )}{3} - \frac{3 sin ^{5} ( y )}{5} + \frac{3 sin ^{7} ( y )}{7} - \frac{sin ^{9} ( y )}{9} + C

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