What is the integral of 1/(sqrt(1-cos(x))) ?

The integral of 1/(sqrt(1-cos(x))) is -sqrt(2)(1/2 ln|sec(x/2)+1|-1/2 ln|sec(x/2)-1|)+C

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integral of 1/(sqrt(1-cos(x)))

\int \frac{1}{1 - cos ( x )} d x

- 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1) + C

Solution steps

\int \frac{1}{1 - cos ( x )} d x

Apply u-substitution

: \int \frac{2}{u 1 + u ^{2}} d u

= \int \frac{2}{u 1 + u ^{2}} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u 1 + u ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2} - 1} d v

= 2 \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 2 \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

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

Substitute back

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

Simplify

2 - \frac{ln 1 + tan ^{2} ( \frac{x}{2} ) + 1}{2} - \frac{ln 1 + tan ^{2} ( \frac{x}{2} ) - 1}{2} : - 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1)

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

Add a constant to the solution

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

x = - 2, x = 2, y = 2 x^{2} + 12, y = 0

\int \frac{x + 1}{x ^{2} + 1} d x

\frac{d}{d x} (8 cosh (\frac{x}{4}) ln (arcsin (x)))

\frac{d ^{2}}{d x ^{2}} (2 x^{- 7})

What is the integral of 1/(sqrt(1-cos(x))) ?
The integral of 1/(sqrt(1-cos(x))) is -sqrt(2)(1/2 ln|sec(x/2)+1|-1/2 ln|sec(x/2)-1|)+C