integral from 0 to 1 of tan^3(x)

解答

\int_{0}^{1} tan^{3} (x) d x

- ln (sec (1)) + \frac{sec ^{2} ( 1 ) - 1}{2}

十进制

0.59713 \dots

求解步骤

\int_{0}^{1} tan^{3} (x) d x

使用三角恒等式改写

= \int_{0}^{1} (- 1 + sec^{2} (x)) tan (x) d x

使用换元积分法

= \int_{1}^{s e c (1)} \frac{- 1 + u ^{2}}{u} d u

乘开

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

= \int_{1}^{s e c (1)} - \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

= - \int_{1}^{s e c (1)} \frac{1}{u} d u + \int_{1}^{s e c (1)} u d u

\int_{1}^{s e c (1)} \frac{1}{u} d u = ln (sec (1))

\int_{1}^{s e c (1)} u d u = \frac{sec ^{2} ( 1 ) - 1}{2}

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

\int_{2}^{5} \frac{1}{x ^{4}} + 10 x^{5} d x

\int_{0}^{1} \frac{e ^{2 x} - 1}{2} d x

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

\int_{0}^{π} - sin^{3} (x) d x

\int_{0}^{L} x \cdot sin (\frac{nπ x}{L}) d x