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受欢迎的微积分 >

integral of tan^3(19pix)

解答

\int tan^{3} (19 π x) d x

解答

\frac{1}{19 π} (- ln ∣ sec (19 π x) ∣ + \frac{sec ^{2} ( 19 π x )}{2}) + C

求解步骤

\int tan^{3} (19 π x) d x

使用换元积分法

= \int tan^{3} (u) \frac{1}{19 π} d u

提出常数:

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

= \frac{1}{19 π} \cdot \int tan^{3} (u) d u

使用三角恒等式改写

= \frac{1}{19 π} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

使用换元积分法

= \frac{1}{19 π} \cdot \int \frac{- 1 + v ^{2}}{v} d v

乘开

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= \frac{1}{19 π} \cdot \int - \frac{1}{v} + v 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

= \frac{1}{19 π} (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{19 π} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

代回

= \frac{1}{19 π} (- ln ∣ sec (19 π x) ∣ + \frac{sec ^{2} ( 19 π x )}{2})

解答补常数

= \frac{1}{19 π} (- ln ∣ sec (19 π x) ∣ + \frac{sec ^{2} ( 19 π x )}{2}) + C

作图

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