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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Apply u-substitution

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

= \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

: 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

= 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

\int e^{x} \cdot cos (2 x) d x

7 x

\int x^{5} e^{x} d x

\frac{d}{d x} (1 x^{2})

\int x^{3} (x^{2} + 1) d x

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