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受欢迎的微积分 >

integral of 31tan^5(x)sec^4(x)

解答

\int 31 tan^{5} (x) sec^{4} (x) d x

解答

31 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

求解步骤

\int 31 tan^{5} (x) sec^{4} (x) d x

提出常数:

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

= 31 \cdot \int tan^{5} (x) sec^{4} (x) d x

使用三角恒等式改写

= 31 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) tan^{5} (x) d x

使用换元积分法

= 31 \cdot \int u^{5} (1 + u^{2}) d u

乘开

u^{5} (1 + u^{2}) : u^{5} + u^{7}

= 31 \cdot \int u^{5} + u^{7} 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

= 31 (\int u^{5} d u + \int u^{7} d u)

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

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

= 31 (\frac{u ^{6}}{6} + \frac{u ^{8}}{8})

u = tan (x)

代回

= 31 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8})

解答补常数

= 31 (\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}) + C

作图

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