integral of tan^6(t)sec^4(t)

解答

\int tan^{6} (t) sec^{4} (t) d t

\frac{tan ^{7} ( t )}{7} + \frac{tan ^{9} ( t )}{9} + C

求解步骤

\int tan^{6} (t) sec^{4} (t) d t

使用三角恒等式改写

= \int (1 + tan^{2} (t)) sec^{2} (t) tan^{6} (t) d t

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 = tan (t)

代回

= \frac{tan ^{7} ( t )}{7} + \frac{tan ^{9} ( t )}{9}

解答补常数

= \frac{tan ^{7} ( t )}{7} + \frac{tan ^{9} ( t )}{9} + C

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