integral from-pi/9 to pi/9 of sec^3(3x)

解答

\int_{- \frac{π}{9}}^{\frac{π}{9}} sec^{3} (3 x) d x

\frac{1}{3} (23 + \frac{1}{2} (ln (3 + 2) - ln (- 3 + 2)))

十进制

1.59368 \dots

求解步骤

\int_{- \frac{π}{9}}^{\frac{π}{9}} sec^{3} (3 x) d x

使用换元积分法

= \int_{- \frac{π}{3}}^{\frac{π}{3}} sec^{3} (u) \frac{1}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int_{- \frac{π}{3}}^{\frac{π}{3}} sec^{3} (u) d u

使用降次积分法

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

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

= \frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{3}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - ln (- 3 + 2)))

化简

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

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

带入上下限计算后得:

23

= \frac{1}{3} (23 + \frac{1}{2} (ln (3 + 2) - ln (- 3 + 2)))

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