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

The integral of 1/(csc(x)-1) is 4(-1/(2(tan(x/2)-1))-1/2 arctan(tan(x/2)))+C

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integral of 1/(csc(x)-1)

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

4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}))) + C

Solution steps

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

Apply u-substitution

: \int \frac{4 u}{( 1 + u ^{2} - 2 u ) ( 1 + u ^{2} )} d u

= \int \frac{4 u}{( 1 + u ^{2} - 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

= 4 \cdot \int \frac{u}{( 1 + u ^{2} - 2 u ) ( 1 + u ^{2} )} d u

Complete the square

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

= 4 \cdot \int \frac{u}{( u - 1 ) ^{2} ( 1 + u ^{2} )} d u

Apply u-substitution

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

= 4 \cdot \int \frac{v + 1}{v ^{2} ( v ^{2} + 2 v + 2 )} d v

Apply u-substitution

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

= 4 \cdot \int \frac{w}{( w - 1 ) ^{2} ( w ^{2} + 1 )} d w

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{2 ( w - 1 ) ^{2}} - \frac{1}{2 ( w ^{2} + 1 )} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{1}{2 ( w - 1 ) ^{2}} d w - \int \frac{1}{2 ( w ^{2} + 1 )} d w)

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

\int \frac{1}{2 ( w ^{2} + 1 )} d w = \frac{1}{2} arctan (w)

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

Substitute back

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

Simplify

4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 + 1 - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}) - 1 + 1)) : 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2})))

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2})))

Add a constant to the solution

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}))) + C

x \to 0 lim (\frac{5 e ^{x} - 5}{x})

\int (cos (x))^{4} d x

\int_{0}^{1} \frac{3}{4 - x ^{2}} d x

f (x) = \frac{x - 3}{x + 4}

\frac{s}{s ^{2} + 3}

What is the integral of 1/(csc(x)-1) ?
The integral of 1/(csc(x)-1) is 4(-1/(2(tan(x/2)-1))-1/2 arctan(tan(x/2)))+C