integral from 0 to pi/4 of 3tan^3(x)

解答

\int_{0}^{\frac{π}{4}} 3 tan^{3} (x) d x

3 (- \frac{1}{2} ln (2) + \frac{1}{2})

十进制

0.46027 \dots

求解步骤

\int_{0}^{\frac{π}{4}} 3 tan^{3} (x) d x

提出常数:

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

= 3 \cdot \int_{0}^{\frac{π}{4}} tan^{3} (x) d x

使用三角恒等式改写

= 3 \cdot \int_{0}^{\frac{π}{4}} (- 1 + sec^{2} (x)) tan (x) d x

使用换元积分法

= 3 \cdot \int_{1}^{2} \frac{- 1 + u ^{2}}{u} d u

乘开

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

= 3 \cdot \int_{1}^{2} - \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

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

\int_{1}^{2} \frac{1}{u} d u = \frac{1}{2} ln (2)

\int_{1}^{2} u d u = \frac{1}{2}

= 3 (- \frac{1}{2} ln (2) + \frac{1}{2})

\frac{\partial}{\partial z} (e^{3 x + 4 y} cos (5 z))

\int (1 + \frac{1}{t})^{3} (\frac{1}{t ^{2}}) d t

n = 1 \sum \infty (0.1)^{n}

\frac{d}{d x} (\frac{2 x + 5}{x ^{2} - 3})

d er i v a t i v e f (x) = - 3 x + 4