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

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

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

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

diverges

Solution steps

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

Factor

cos (x) - 1 : - (- cos (x) + 1)

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

Take the constant out:

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

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

Apply u-substitution

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

= - \int_{1}^{- 1} - \frac{1}{( - u + 1 ) 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}{( - u + 1 ) 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}{( - u + 1 ) 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_{0}^{3} \frac{1}{x ^{2} + 1} d x

\int_{- 1}^{4} 5 x^{4} - 8 x^{3} + 6 d x

\int_{0}^{x} cosh^{2} (t) d t

\int_{0}^{x} \frac{1}{1 + t + t ^{2}} d t

\int_{0}^{2} \frac{1}{( x - 2 ) ^{\frac{3}{5}}} d x

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