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

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

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

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

Apply u-substitution

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

Complete the square

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

= 4 (- \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 (- \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

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