integral of tan^4(9x)

解答

\int tan^{4} (9 x) d x

\frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x) + C

求解步骤

\int tan^{4} (9 x) d x

使用换元积分法

= \int tan^{4} (u) \frac{1}{9} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int tan^{4} (u) d u

使用降次积分法

= \frac{1}{9} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

= \frac{1}{9} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

u = 9 x

代回

= \frac{1}{9} (\frac{tan ^{3} ( 9 x )}{3} - (- 9 x + tan (9 x)))

化简

\frac{1}{9} (\frac{tan ^{3} ( 9 x )}{3} - (- 9 x + tan (9 x))) : \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x)

= \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x)

解答补常数

= \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x) + C

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