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受欢迎的微积分 >

integral from 0 to pi of 1/(1-cos(x))

解答

\int_{0}^{π} \frac{1}{1 - cos ( x )} d x

解答

发散

求解步骤

\int_{0}^{π} \frac{1}{1 - cos ( x )} d x

使用换元积分法

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

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

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

提出常数:

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

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

使用换元积分法

= - (- \int_{2}^{0} - \frac{1}{v - v ^{2} + 2 v} d v)

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

= - (- (- \int_{0}^{2} - \frac{1}{v - v ^{2} + 2 v} d v))

提出常数:

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

= - (- (- (- \int_{0}^{2} \frac{1}{v - v ^{2} + 2 v} d v)))

对

- v^{2} + 2 v

配方:

- (v - 1)^{2} + 1

= - - - - \int_{0}^{2} \frac{1}{v - ( v - 1 ) ^{2} + 1} d v

使用换元积分法

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

使用三角换元法

= - (- (- (- \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{sin ( t ) + 1} d t)))

使用换元积分法

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

提出常数:

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

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

对

2 r + 1 + r^{2}

配方:

(r + 1)^{2}

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

使用换元积分法

= - (- (- (- 2 \cdot \int_{0}^{2} \frac{1}{s ^{2}} d s)))

使用幂法则

= - (- (- (- 2 [- \frac{1}{s}]_{0}^{2})))

化简

= 2 [- \frac{1}{s}]_{0}^{2}

带入上下限计算后得:

发散

= 发散

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\int_{1}^{2} 2 x e^{x} - 1 d x

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\int_{1}^{3} t^{2} d t

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\int_{1}^{2} (\frac{x ^{2} + 3}{x}) d x

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\int_{0}^{9} \frac{10 x}{( 1 + x ) ^{2}} d x