integral of sec^4(3x)tan^3(3x)

解答

\int sec^{4} (3 x) tan^{3} (3 x) d x

\frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12} + C

求解步骤

\int sec^{4} (3 x) tan^{3} (3 x) d x

使用三角恒等式改写

= \int (- 1 + sec^{2} (3 x)) tan (3 x) sec^{4} (3 x) d x

使用换元积分法

= \int \frac{u ^{3} ( u ^{2} - 1 )}{3} d u

乘开

\frac{u ^{3} ( u ^{2} - 1 )}{3} : \frac{u ^{5}}{3} - \frac{u ^{3}}{3}

= \int \frac{u ^{5}}{3} - \frac{u ^{3}}{3} 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 ^{5}}{3} d u - \int \frac{u ^{3}}{3} d u

\int \frac{u ^{5}}{3} d u = \frac{u ^{6}}{18}

\int \frac{u ^{3}}{3} d u = \frac{u ^{4}}{12}

= \frac{u ^{6}}{18} - \frac{u ^{4}}{12}

u = sec (3 x)

代回

= \frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12}

解答补常数

= \frac{sec ^{6} ( 3 x )}{18} - \frac{sec ^{4} ( 3 x )}{12} + C

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