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受欢迎的微积分 >

integral from 0 to pi/4 of sqrt(tan(x))

解答

\int_{0}^{\frac{π}{4}} tan (x) d x

解答

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

+1

十进制

0.48749 \dots

求解步骤

\int_{0}^{\frac{π}{4}} tan (x) d x

使用三角换元法

= \int_{0}^{1} \frac{u}{1 + u ^{2}} d u

使用换元积分法

= \int_{0}^{2} \frac{2 v}{4 + v ^{2}} d v

提出常数:

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

= 2 \cdot \int_{0}^{2} \frac{v}{4 + v ^{2}} d v

使用换元积分法

= 2 \cdot \int_{0}^{2} \frac{2 w ^{2}}{4 + w ^{4}} d w

提出常数:

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

= 2 \cdot 2 \cdot \int_{0}^{2} \frac{w ^{2}}{4 + w ^{4}} d w

将

\frac{w ^{2}}{4 + w ^{4}}

用部份分式展开:

- \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )}

= 2 \cdot 2 \cdot \int_{0}^{2} - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w

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 \cdot 2 (- \int_{0}^{2} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w + \int_{0}^{2} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w)

\int_{0}^{2} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w = \frac{1}{4} (\frac{1}{2} ln (2 + 2) - arctan (2 + 1) + \frac{π}{4})

\int_{0}^{2} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w = \frac{1}{4} (\frac{1}{2} ln (2 - 2) + arctan (2 - 1) + \frac{π}{4})

= 2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (2 + 2) - arctan (2 + 1) + \frac{π}{4}) + \frac{1}{4} (\frac{1}{2} ln (2 - 2) + arctan (2 - 1) + \frac{π}{4}))

化简

2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (2 + 2) - arctan (2 + 1) + \frac{π}{4}) + \frac{1}{4} (\frac{1}{2} ln (2 - 2) + arctan (2 - 1) + \frac{π}{4})) : \frac{2 ln ( 2 - 2 ) + 4 arctan ( 1 + 2 ) + 4 arctan ( 2 - 1 ) - 2 ln ( 2 + 2 )}{4 2}

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

作图

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