integral of tan^3(2x)sec^6(2x)

解答

\int tan^{3} (2 x) sec^{6} (2 x) d x

\frac{sec ^{8} ( 2 x )}{16} - \frac{sec ^{6} ( 2 x )}{12} + C

求解步骤

\int tan^{3} (2 x) sec^{6} (2 x) d x

使用三角恒等式改写

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

使用换元积分法

= \int \frac{u ^{5} ( u ^{2} - 1 )}{2} d u

乘开

\frac{u ^{5} ( u ^{2} - 1 )}{2} : \frac{u ^{7}}{2} - \frac{u ^{5}}{2}

= \int \frac{u ^{7}}{2} - \frac{u ^{5}}{2} 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 \frac{u ^{7}}{2} d u - \int \frac{u ^{5}}{2} d u

\int \frac{u ^{7}}{2} d u = \frac{u ^{8}}{16}

\int \frac{u ^{5}}{2} d u = \frac{u ^{6}}{12}

= \frac{u ^{8}}{16} - \frac{u ^{6}}{12}

u = sec (2 x)

代回

= \frac{sec ^{8} ( 2 x )}{16} - \frac{sec ^{6} ( 2 x )}{12}

解答补常数

= \frac{sec ^{8} ( 2 x )}{16} - \frac{sec ^{6} ( 2 x )}{12} + C

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