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受欢迎的微积分 >

integral of 2tan^4(x)sec^6(x)

解答

\int 2 tan^{4} (x) sec^{6} (x) d x

解答

2 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

求解步骤

\int 2 tan^{4} (x) sec^{6} (x) d x

提出常数:

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

= 2 \cdot \int tan^{4} (x) sec^{6} (x) d x

使用三角恒等式改写

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

使用换元积分法

= 2 \cdot \int u^{4} (1 + u^{2})^{2} d u

乘开

u^{4} (1 + u^{2})^{2} : u^{4} + 2 u^{6} + u^{8}

= 2 \cdot \int u^{4} + 2 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

= 2 (\int u^{4} d u + \int 2 u^{6} d u + \int u^{8} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 2 (\frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9})

u = tan (x)

代回

= 2 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9})

解答补常数

= 2 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

作图

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