integral from-pi/6 to 0 of 4sec^3(x)

解答

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

\frac{4}{3} + ln (3)

十进制

2.43194 \dots

求解步骤

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

提出常数:

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

= 4 \cdot \int_{- \frac{π}{6}}^{0} sec^{3} (x) d x

使用降次积分法

= 4 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \int_{- \frac{π}{6}}^{0} sec (x) d x)

\int_{- \frac{π}{6}}^{0} sec (x) d x = \frac{1}{2} ln (3)

= 4 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

化简

4 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{6}}^{0} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 4 [\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + ln (3)

= 4 [\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{6}}^{0} + ln (3)

带入上下限计算后得:

\frac{1}{3}

= 4 \cdot \frac{1}{3} + ln (3)

化简

= \frac{4}{3} + ln (3)

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