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

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

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integral of 1/(csc(x)(1+sin(x)))

\int \frac{1}{csc ( x ) ( 1 + sin ( x ) )} d x

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

Solution steps

\int \frac{1}{csc ( x ) ( 1 + sin ( x ) )} d x

Rewrite using trig identities

= \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

Apply Integration By Parts

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

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

= 4 (u (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2 ( u + 1 )} - \frac{1}{4} ln u^{2} + 1) - (\frac{1}{2} u ln ∣ u + 1 ∣ - \frac{u}{2 ( u + 1 )} - \frac{1}{4} u ln u^{2} + 1 - \frac{1}{2 ( u + 1 )} - \frac{1}{2} arctan (u)))

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

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