integral of sin^6(x)*cos^3(x)

解答

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

\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

求解步骤

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

使用三角恒等式改写

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

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 = sin (x)

代回

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}

解答补常数

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

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