Question

How do you find the ingral of=
ʃ(Chx)^2 dx

Answer 1

What are C and h? Constants?

Answer 2

CHx it's a hyperbolic function.

Answer 3

you mean\[\int \cosh^2 xdx\]?

Answer 4

hyperbolic cosine function, yes?

Answer 5

\[\int\limits_{}^{} (Chx )^{2} dx\]

Answer 6

yes

Answer 7

Chx = Coshx = hyperbolic cosine function

Answer 8

use the definition\[\cosh x={e^x+e^{-x}\over2}\]

Answer 9

\[\frac{ 1 }{ 4 } \int\limits_{}^{} (e ^{x} + e ^{-x})^{2} dx\]

Answer 10

like this?

Answer 11

\[\frac{1}{4}\int\limits_{}^{} (e ^{2x} + e ^{-2x} + 2 ) dx\]

Answer 12

Then how should I proceed?

Answer 13

Integrate term-by-term.

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @naylah
\[\frac{1}{4}\int\limits_{}^{} (e ^{2x} + e ^{-2x} + 2 ) dx=\frac{1}{4}\int e ^{2x}~dx + \frac{1}{4}\int e ^{-2x}~dx + \frac{1}{2}\int dx\]
\(\color{blue}{\text{End of Quote}}\)

Answer 15

\[\frac{ 1 }{ 4 } (\frac{ e ^{2x} }{ 2 }) + \frac{ 1 }{ 4 } (\frac{ e ^{-2x} }{ -2 }) + \frac{ x }{ 2 }\]

Answer 16

\[\frac{ 1 }{ 2 } (\frac{ e ^{2x} - e ^{-2x} }{ 4 }) + \frac{ x }{ 2 } + c\]

Answer 17

\[\frac{ (Shx) ^{2} }{ 2 } + \frac{ x }{ 2 } + c\]

Answer 18

I did as ypu told me but on the book the result is different. On the book it says that the result is: \[\frac{ (ShxChx + x) }{ 2 } + c\]

Answer 19

\[\Large\bf\sf \left(\frac{e^{2x}-e^{-2x}}{4}\right)\quad\ne\quad \frac{(Shx)^2}{2}\]

\[\Large\bf\sf (Shx)^2\quad=\quad \left(\frac{e^x-e^{-x}}{2}\right)^2\quad=\quad \frac{1}{4}\left(e^{2x}-2+e^{-2x}\right)\]
We would need that extra middle term to simplify it down to Shx, yes?

Answer 20

Oh I guess it might be better to do something like this.\[\Large\bf\sf \frac{1}{4}\left(\frac{e^{2x}-e^{-2x}}{2}\right)\quad=\quad \frac{1}{4}\sinh(2x)\quad=\quad \frac{1}{2}Shx \cdot Chx\]Applying Double Angle Formula.

Answer 21

Sorry for my mistake and thank you :D