integral from 0 to pi/4 of sec^5(x)

解答

\int_{0}^{\frac{π}{4}} sec^{5} (x) d x

\frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

十进制

1.56795 \dots

求解步骤

\int_{0}^{\frac{π}{4}} sec^{5} (x) d x

使用降次积分法

= [\frac{sec ^{4} ( x ) sin ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{4}} sec^{3} (x) d x

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

= [\frac{sec ^{4} ( x ) sin ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{2 + ln ( 1 + 2 )}{2}

化简

[\frac{sec ^{4} ( x ) sin ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{2 + ln ( 1 + 2 )}{2} : [\frac{1}{4} sec^{4} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{3 ( 2 + ln ( 1 + 2 ) )}{8}

= [\frac{1}{4} sec^{4} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{3 ( 2 + ln ( 1 + 2 ) )}{8}

带入上下限计算后得:

\frac{1}{2}

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

化简

= \frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

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