integral from 0 to pi/3 of 8sec^3(x)

解答

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

8 (3 + \frac{1}{2} ln (2 + 3))

十进制

19.12423 \dots

求解步骤

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

提出常数:

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

= 8 \cdot \int_{0}^{\frac{π}{3}} sec^{3} (x) d x

使用降次积分法

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} sec (x) d x)

\int_{0}^{\frac{π}{3}} sec (x) d x = ln (2 + 3)

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

化简

8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3)) : 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

= 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{3}} + \frac{1}{2} ln (2 + 3))

带入上下限计算后得:

3

= 8 (3 + \frac{1}{2} ln (2 + 3))

t an g e n t f (x) = 7 x + 2, (2, 4)

x y^{^{'}} + y = e^{x}, y (1) = 1

s l o p e in t erce pt (3, 6), (5, - 8)

\frac{\partial}{\partial x} (ln (x + t^{2}))

\int_{- 2}^{- 1} x f (x^{2} + 1) d x