integral from-1 to 1 of sec^3(x)

解答

\int_{- 1}^{1} sec^{3} (x) d x

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

十进制

4.10866 \dots

求解步骤

\int_{- 1}^{1} sec^{3} (x) d x

使用降次积分法

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

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

= [\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- 1}^{1} + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1)))

化简

[\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- 1}^{1} + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1))) : [\frac{1}{2} sec (x) tan (x)]_{- 1}^{1} + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1)))

= [\frac{1}{2} sec (x) tan (x)]_{- 1}^{1} + \frac{1}{2} (ln (tan (1) + sec (1)) - ln (- tan (1) + sec (1)))

带入上下限计算后得:

sec (1) tan (1)

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

t an g e n t f (x) = \frac{( 2 x + 1 )}{x + 2}, at x = 1

\frac{\partial}{\partial x} (x^{6} y^{6} + 9 x^{4} y)

s l o p e y = 2 x^{2} - 3 x, (2, 2)

x \to 2 - lim (\frac{x + 2}{x - 2})

d er i v a t i v e y = x - 2