Question

integrate (sinhx + 2 coshx)/(sinhx +coshx)

Answer 1

should i let u= sinhx +coshx ?

Answer 2

sinh + cosh = ??

Answer 3

1??

Answer 4

multiple both numerator and denominator by (cosh (x) -sinh (x))

Answer 5

Hints


\[\Large \sinh(x) = \frac{e^x-e^{-x}}{2}\]
\[\Large \cosh(x) = \frac{e^x+e^{-x}}{2}\]


http://math2.org/math/trig/hyperbolics.htm

Answer 6

Then denominator = \(cosh^2 (x) - sinh^2(x) =1\)
you have just numerator, right? What do you have?

Answer 7

wait a moment please

Answer 8

@Loser66 the numerator = \[2\cosh^2x -\sinh^2x -coshxsinhx\]

Answer 9

yup, and 2 cosh^2 = cosh^2 + cosh^2 , right?
again, cosh^2 - sinh^2 =1,
Now, what do you get?

Answer 10

(sinh^2 + 1)(sinh^2+1)

Answer 11

Use these identities

\[\Large {\color{red}{\sinh(x)}} = {\color{red}{\frac{e^x-e^{-x}}{2}}}\]
\[\Large {\color{blue}{\cosh(x)}} = {\color{blue}{\frac{e^x+e^{-x}}{2}}}\]


To make these substitutions


\[\Large \frac{\sinh(x)+2\cosh(x)}{\sinh(x)+\cosh(x)}\]


\[\Large \frac{{\color{red}{\sinh(x)}}+2*{\color{blue}{\cosh(x)}}}{{\color{red}{\sinh(x)}}+{\color{blue}{\cosh(x)}}}\]


\[\Large \frac{{\color{red}{\frac{e^x-e^{-x}}{2}}}+2*{\color{blue}{\frac{e^x+e^{-x}}{2}}}}{{\color{red}{\frac{e^x-e^{-x}}{2}}}+{\color{blue}{\frac{e^x+e^{-x}}{2}}}}\]


Now simplify. After simplifying, the integration should be much easier.

Answer 12

If you want to follow Mr. JIm's way, it's ok. There are many ways to get the answer. I finish my way :)
\(2 cosh^2 (x) -sinh^2 (x) -cosh(x) sinh(x) \\= cosh^2(x) +cosh^2(x) -sinh^2(x) -cosh(x)sinh(x)\\=cosh^2(x) -1 -cosh(x) sinh(x)\)
Now take integral term by term. 1 and cosh(x) sinh(x) are easy, right?
Just the first element. We have \(cosh(2x) = 2cosh^2(x) -1\), then \(cosh^2 (x) = cosh(2x)/2+1/2\)
Every thing is done. right? Of course, you have to put them in neat. Ok?

Answer 13

why 1 is negative, arent cosh^2 x=1+sinh^2 x

Answer 14

yes, my bad :)

Answer 15

you have to be careful there, when applying the same rules of reg trig functions to hyperbolic...

Answer 16

\(2\cosh^2(x)-1=\cosh(2x)\)
is not an identityy...

Answer 17

@SolomonZelman Yes, I read it before using it https://en.wikipedia.org/wiki/Hyperbolic_function

Answer 18

omg, thatlink is so nice... tnx 66 !

Answer 19

yeah, it is an identity, my badd

Answer 20

purposefully

Answer 21

well, if you want,
Loser66 :) ... why 66, btw?

Answer 22

Age changes, username on os doesn't ... anyway ... that is reasonable however...

Answer 23

back to problem, won't distract...

Answer 24

eunna, did you simplify as directed by Jim Thompson?

Answer 25

i try to simplify, but im bad at simplifying, so the answer is weird.. :')

Answer 26

\({\LARGE \displaystyle \frac{ \frac{e^x-e^{-x}}{2} + 2\times \frac{e^x+e^{-x}}{2} }{ \frac{e^x-e^{-x}}{2}+ \frac{e^x+e^{-x}}{2} }}\)

\({\LARGE \displaystyle \frac{ \frac{e^x-e^{-x}}{2} + \frac{2e^x+2e^{-x}}{2} }{ \frac{e^x-e^{-x}}{2}+ \frac{e^x+e^{-x}}{2} }}\)

Answer 27

\({\LARGE \displaystyle \frac{ \frac{e^x-e^{-x}+2e^x+2e^{-x}}{2} }{ \frac{e^x-e^{-x}+e^x+e^{-x}}{2} }}\)

Answer 28

show me please, what do you get after simplifying this...

Answer 29

you can multiply times two on top and bottom for simplicity

\({\Large \displaystyle \frac{ e^x-e^{-x}+2e^x+2e^{-x} }{e^x-e^{-x}+e^x+e^{-x} }}\)

Answer 30

please continue

Answer 31

give me a moment

Answer 32

k

Answer 33

\[\frac{ e^x -e ^{-x}+2e^x+2e-x}{ \frac{ 2 }{ e^x } }\]

Answer 34

when yo add everything on top and bottom,

\({\Large \displaystyle \frac{ 3e^x+e^{-x}}{2e^x }}\)

Answer 35

\({\Large \displaystyle \frac{ 3e^x}{2e^x }+\frac{ e^{-x}}{2e^x }}\)

\({\Large \displaystyle \frac{ 3}{2 }+\frac{ 1}{2e^{2x} }}\)

Answer 36

so really that is what you are integrating ... (shouldn't be so hard to do)

Answer 37

i got different answer for both method

Answer 38

thank you guys :)