integral of sec^{10}(6x)tan^3(6x)

解答

\int sec^{10} (6 x) tan^{3} (6 x) d x

\frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60} + C

求解步骤

\int sec^{10} (6 x) tan^{3} (6 x) d x

使用三角恒等式改写

= \int (- 1 + sec^{2} (6 x)) tan (6 x) sec^{10} (6 x) d x

使用换元积分法

= \int \frac{u ^{9} ( u ^{2} - 1 )}{6} d u

乘开

\frac{u ^{9} ( u ^{2} - 1 )}{6} : \frac{u ^{11}}{6} - \frac{u ^{9}}{6}

= \int \frac{u ^{11}}{6} - \frac{u ^{9}}{6} 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 ^{11}}{6} d u - \int \frac{u ^{9}}{6} d u

\int \frac{u ^{11}}{6} d u = \frac{u ^{12}}{72}

\int \frac{u ^{9}}{6} d u = \frac{u ^{10}}{60}

= \frac{u ^{12}}{72} - \frac{u ^{10}}{60}

u = sec (6 x)

代回

= \frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60}

解答补常数

= \frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60} + C

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