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受欢迎的微积分 >

integral of tan^{11}(x)sec^8(x)

解答

\int tan^{11} (x) sec^{8} (x) d x

解答

\frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

求解步骤

\int tan^{11} (x) sec^{8} (x) d x

使用三角恒等式改写

= \int sec^{2} (x) (1 + tan^{2} (x))^{3} tan^{11} (x) d x

使用换元积分法

= \int u^{11} (1 + u^{2})^{3} d u

乘开

u^{11} (1 + u^{2})^{3} : u^{11} + 3 u^{13} + 3 u^{15} + u^{17}

= \int u^{11} + 3 u^{13} + 3 u^{15} + u^{17} 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^{11} d u + \int 3 u^{13} d u + \int 3 u^{15} d u + \int u^{17} d u

\int u^{11} d u = \frac{u ^{12}}{12}

\int 3 u^{13} d u = \frac{3 u ^{14}}{14}

\int 3 u^{15} d u = \frac{3 u ^{16}}{16}

\int u^{17} d u = \frac{u ^{18}}{18}

= \frac{u ^{12}}{12} + \frac{3 u ^{14}}{14} + \frac{3 u ^{16}}{16} + \frac{u ^{18}}{18}

u = tan (x)

代回

= \frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18}

解答补常数

= \frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

作图

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