What is the integral of e^xcsc(e^x+1)cot(e^x+1) ?

The integral of e^xcsc(e^x+1)cot(e^x+1) is -csc(e^x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^xcsc(e^x+1)cot(e^x+1)

\int e^{x} csc (e^{x} + 1) cot (e^{x} + 1) d x

- csc (e^{x} + 1) + C

Solution steps

\int e^{x} csc (e^{x} + 1) cot (e^{x} + 1) d x

Apply u-substitution

: \int csc (u) cot (u) d u

= \int csc (u) cot (u) d u

Apply u-substitution

: \int - 1 d v

= \int - 1 d v

Integral of a constant:

\int a d x = a x

= (- 1) v

Substitute back

= (- 1) csc (e^{x} + 1)

Simplify

= - csc (e^{x} + 1)

Add a constant to the solution

= - csc (e^{x} + 1) + C

\int 2 - 4 x d x

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

\int sin (x) + 4 d x

\int (x^{2} - 1) csc^{3} (2 x^{3} - 6 x) d x

\int \frac{- 7 x}{( x + 1 ) - x + 1} d x

What is the integral of e^xcsc(e^x+1)cot(e^x+1) ?
The integral of e^xcsc(e^x+1)cot(e^x+1) is -csc(e^x+1)+C