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受欢迎的微积分 >

integral of (tan(x/(18)))^5

解答

\int (tan (\frac{x}{18}))^{5} d x

解答

18 (ln sec (\frac{x}{18}) - sec^{2} (\frac{x}{18}) + \frac{sec ^{4} ( \frac{x}{18} )}{4}) + C

求解步骤

\int (tan (\frac{x}{18}))^{5} d x

化简

= \int tan^{5} (\frac{x}{18}) d x

使用换元积分法

= \int tan^{5} (u) \cdot 18 d u

提出常数:

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

= 18 \cdot \int tan^{5} (u) d u

使用三角恒等式改写

= 18 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) d u

使用换元积分法

= 18 \cdot \int \frac{( - 1 + v ^{2} ) ^{2}}{v} d v

乘开

\frac{( - 1 + v ^{2} ) ^{2}}{v} : \frac{1}{v} - 2 v + v^{3}

= 18 \cdot \int \frac{1}{v} - 2 v + v^{3} d v

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

= 18 (\int \frac{1}{v} d v - \int 2 v d v + \int v^{3} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int 2 v d v = v^{2}

\int v^{3} d v = \frac{v ^{4}}{4}

= 18 (ln ∣ v ∣ - v^{2} + \frac{v ^{4}}{4})

代回

= 18 (ln sec (\frac{x}{18}) - sec^{2} (\frac{x}{18}) + \frac{sec ^{4} ( \frac{x}{18} )}{4})

解答补常数

= 18 (ln sec (\frac{x}{18}) - sec^{2} (\frac{x}{18}) + \frac{sec ^{4} ( \frac{x}{18} )}{4}) + C

作图

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