What is the integral of 1/(csc(2x)-1) ?

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

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integral of 1/(csc(2x)-1)

\int \frac{1}{csc ( 2 x ) - 1} d x

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

Solution steps

\int \frac{1}{csc ( 2 x ) - 1} d x

Apply u-substitution

: \int \frac{1}{2 ( csc ( u ) - 1 )} d u

= \int \frac{1}{2 ( csc ( u ) - 1 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{csc ( u ) - 1} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

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

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \cdot e^{- x + 1}

\frac{\partial}{\partial x} (e^{(- x^{2} - y^{2})})

\frac{d}{d x} ((5 x^{2} + x + 9) tan (7 x))

\frac{\partial}{\partial y} (6 x y^{2} + x^{2})

x \to 5 - lim (\frac{- 2}{x ^{2} - 25})

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