integral of cos^6(y)sin^3(y)

解答

\int cos^{6} (y) sin^{3} (y) d y

- \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9} + C

求解步骤

\int cos^{6} (y) sin^{3} (y) d y

使用三角恒等式改写

= \int (1 - cos^{2} (y)) sin (y) cos^{6} (y) d y

使用换元积分法

= \int - u^{6} (1 - u^{2}) d u

乘开

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

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

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

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

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

u = cos (y)

代回

= - \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9}

解答补常数

= - \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9} + C

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