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受欢迎的微积分 >

integral from 0 to pi/(12) of 6tan(3x)

解答

\int_{0}^{\frac{π}{12}} 6 tan (3 x) d x

解答

ln (2)

+1

十进制

0.69314 \dots

求解步骤

\int_{0}^{\frac{π}{12}} 6 tan (3 x) d x

提出常数:

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

= 6 \cdot \int_{0}^{\frac{π}{12}} tan (3 x) d x

使用三角恒等式改写

= 6 \cdot \int_{0}^{\frac{π}{12}} \frac{sin ( 3 x )}{cos ( 3 x )} d x

使用换元积分法

= 6 \cdot \int_{1}^{\frac{2}{2}} - \frac{1}{3 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 6 (- \int_{\frac{2}{2}}^{1} - \frac{1}{3 u} d u)

提出常数:

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

= 6 (- (- \frac{1}{3} \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 (- (- \frac{1}{3} [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

化简

6 (- (- \frac{1}{3} [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 2 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 2 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

带入上下限计算后得:

\frac{1}{2} ln (2)

= 2 \cdot \frac{1}{2} ln (2)

化简

= ln (2)

作图

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