How do you integrate $\int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x$ $int (e^x+e^-x)/(e^x-e^-x)dx$ ?

Question

How do you integrate $\int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x$ $int (e^x+e^-x)/(e^x-e^-x)dx$ ?

1 Answer

Impact of this question

5625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T02:39:45

$\int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x = ln | sinh x | + C = ln ∣ ∣ ∣ \frac{1}{2} (e^{x} - e^{- x}) ∣ ∣ ∣ + C$ $int (e^x + e^-x)/(e^x - e^-x)dx = ln|sinhx| + C = ln|1/2(e^x -e^-x)| + C$

Explanation:

We know that

$∙ cosh x = \frac{e^{x} + e^{- x}}{2}$ $•coshx = (e^x + e^-x)/2$
$∙ sinh x = \frac{e^{x} - e^{- x}}{2}$ $•sinhx = (e^x- e^-x)/2$

This integral can be rewritten as

$\int \frac{cosh x}{sinh x} d x$ $int coshx/sinhx dx$ , where $cosh x$ $coshx$ and $sinh x$ $sinhx$ represent the hyperbolic trigonometric functions

Now use a substitution to solve. Let $u = sinh x$ $u = sinhx$ . Just like with regular trigonometric functions, $d u = cosh x d x$ $du = coshx dx$ and $d x = \frac{d u}{cosh x}$ $dx= (du)/coshx$ .

$\int \frac{cosh x}{u} \cdot \frac{d u}{cosh x}$ $int coshx/u * (du)/coshx$

$\int \frac{1}{u} d u$ $int 1/u du$

$ln | u | + C$ $ln|u| + C$

$ln | sinh x | + C$ $ln|sinhx| + C$

If you wish, the answer can be written as

$ln ∣ ∣ ∣ \frac{1}{2} (e^{x} - e^{- x}) ∣ ∣ ∣ + C$ $ln|1/2(e^x - e^-x)| + C$

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x$ $int (e^x+e^-x)/(e^x-e^-x)dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int (e^x+e^-x)/(e^x-e^-x)dx∫ex+e−xex−e−xdxint (e^x+e^-x)/(e^x-e^-x)dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}} d x$ $int (e^x+e^-x)/(e^x-e^-x)dx$ ?