Question

Evaluate the intergal

Answer 1

\[\pi \int\limits_{0}^{1}((e^x+e ^{-x})\sqrt{1+\frac{ 1 }{ 4 }(e^x-e ^{-x})^2})dx\]

Answer 2

If it wasn't for that damn 1 in the square root, this would be simple.

Answer 3

you tried substituting u =e^x - e^(-x) , did you ?

Answer 4

Why would I do that?

Answer 5

because ,then (e^x+e^(-x))dx = du.
and your integral reduces to \(\int \sqrt{1+(1/4)u^2}du\)
with limits changed.

Answer 6

Yep, Just saw it.

Answer 7

Seriously @hartnn , how do you come up with all this stuff? :P .

Answer 8

Practice, lots of it. Then it automatically comes in mind,

Answer 9

Well the new limits would be 0 and e-(1/e) right?

Answer 10

yes.

Answer 11

Am I correct in saying I see a trig sub now?

Answer 12

I would set u = 2tan(theta) right?

Answer 13

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

Answer 14

Well... We haven't learn't integrals of hyperbolic trig functions.

Answer 15

Have you learned any hyperbolic identities?

Answer 16

\(\\ \large 22. \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln| x+\sqrt{x^2+a^2}|+c
\\\)

Answer 17

Yeah we have. But I would rather avoid hyperobolics.

Answer 18

\[
1+\sinh^2(x) = \cosh^2(x)
\]This would clear out your square root.

Answer 19

Hmm...

Answer 20

You could always convert back to \(\exp(\;)\) when you are done.

Answer 21

What about the x^x-e^-x then?

Answer 22

Inside the square root.

Answer 23

And thanks Hartann! I got it :) . Just trying to see this alternate method.

Answer 24

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

Answer 25

Ohh that's just beautiful.

Answer 26

\[
\sqrt{1+\sinh^2(x)}=\sqrt{\cosh^2x)} = |\cosh(x)|=\cosh(x)
\]

Answer 27

ofcourse! @wio method is way better than mine! :)

Answer 28

That's so beautiful it brought a tear to my eye.

Answer 29

\[
\pi\int_0^1 2\cosh^2(x)\;dx
\]

Answer 30

KK so now I have:\[2\pi \int\limits_{0}^{1}\cosh^2(x) dx\]

Answer 31

Use half angles right?

Answer 32

Yep.

Answer 33

\[
2\pi\int_0^1\frac{\cosh(2x)+1}{2}dx
\]

Answer 34

Yep, I got that.

Answer 35

Then integrate them separately and use \(u=2x\) sub.

Answer 36

\[
(\sinh(x))' = \cosh(x)
\]

Answer 37

lololol

Answer 38

Go away.

Answer 39

HEY!

Answer 40

Thanks a lot wio! Got it :) .

Answer 41

i came here to give you my assistance in U SUBSTITUTIOUn

Answer 42

lmao it bought a tear to your eye!

Answer 43

There is no section of hyperbolic integrals in many calculus classes.

Answer 44

subhashis are u andra pradesh side by any chance

Answer 45

No.

Answer 46

where are u from

Answer 47

hartnn will i get banned from too many warnings

Answer 48

Do not spam on questions, if you want to ask anything that is unrelated to questions, use personal message facility.

Answer 49

because this is my 2nd warning in 2 days

Answer 50

if you continue such behaviour, you'll get suspended.

Answer 51

what i actually drew was a x^4 graph

Answer 52

lmao :)

Answer 53

Stop commenting right there!

Answer 54

it was a x^4 graph with cuspid inflection points obbvioussly