integral of sec^5(x)tan^5(x)

解答

\int sec^{5} (x) tan^{5} (x) d x

\frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5} + C

求解步骤

\int sec^{5} (x) tan^{5} (x) d x

使用三角恒等式改写

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{5} (x) d x

使用换元积分法

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

乘开

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

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

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= \frac{u ^{9}}{9} - \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}

u = sec (x)

代回

= \frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5}

解答补常数

= \frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5} + C