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受欢迎的微积分 >

integral of 9tan^5(x)

解答

\int 9 tan^{5} (x) d x

解答

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

求解步骤

\int 9 tan^{5} (x) d x

提出常数:

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

= 9 \cdot \int tan^{5} (x) d x

使用三角恒等式改写

= 9 \cdot \int (- 1 + sec^{2} (x))^{2} tan (x) d x

使用换元积分法

= 9 \cdot \int \frac{( - 1 + u ^{2} ) ^{2}}{u} d u

乘开

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

= 9 \cdot \int \frac{1}{u} - 2 u + u^{3} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

= 9 (ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4})

u = sec (x)

代回

= 9 (ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4})

解答补常数

= 9 (ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}) + C

作图

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