What is the integral from 0 to pi of 1/(1-cos(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Apply u-substitution

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

= \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

Take the constant out:

\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)

Apply u-substitution

: \int_{2}^{0} - \frac{1}{v - v ^{2} + 2 v} d v

= - (- \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))

Take the constant out:

\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)))

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

: \int_{- 1}^{1} \frac{2}{2 r + 1 + r ^{2}} d r

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

Take the constant out:

\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)))

Complete the square

2 r + 1 + r^{2} : (r + 1)^{2}

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

Apply u-substitution

: \int_{0}^{2} \frac{1}{s ^{2}} d s

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{1} x 16 x^{2} + 9 d x

\int_{6}^{8} 4 - 10 e^{- 0.2 x} d x

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

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

What is the integral from 0 to pi of 1/(1-cos(x)) ?
The integral from 0 to pi of 1/(1-cos(x)) is diverges