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受欢迎的微积分 >

integral from 0 to pi/4 of 2tan^5(x)

解答

\int_{0}^{\frac{π}{4}} 2 tan^{5} (x) d x

解答

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

+1

十进制

0.19314 \dots

求解步骤

\int_{0}^{\frac{π}{4}} 2 tan^{5} (x) d x

提出常数:

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

= 2 \cdot \int_{0}^{\frac{π}{4}} tan^{5} (x) d x

使用三角恒等式改写

= 2 \cdot \int_{0}^{\frac{π}{4}} (- 1 + sec^{2} (x))^{2} tan (x) d x

使用换元积分法

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

乘开

\frac{( - 1 + u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u + u^{3}

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{2} \frac{1}{u} d u = \frac{1}{2} ln (2)

\int_{1}^{2} 2 u d u = 1

\int_{1}^{2} u^{3} d u = \frac{3}{4}

= 2 (\frac{1}{2} ln (2) - 1 + \frac{3}{4})

化简

2 (\frac{1}{2} ln (2) - 1 + \frac{3}{4}) : - \frac{1}{2} + ln (2)

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

作图

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