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受欢迎的微积分 >

integral of 9tan^3(θ)

解答

\int 9 tan^{3} (θ) d θ

解答

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

求解步骤

\int 9 tan^{3} (θ) d θ

提出常数:

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

= 9 \cdot \int tan^{3} (θ) d θ

使用三角恒等式改写

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

使用换元积分法

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

乘开

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 9 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

u = sec (θ)

代回

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

解答补常数

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

作图

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