Transformation of arctan(sinh(x)) ??

khaixiang

5

It confounds me on how to change from $\arctan(\sinh(x))$ to $2\arctan(e^{x})-\frac{\pi}{2}$ Can anyone help?

Reply 1

dvs

10

I can think of a way, but it's very dirty.

Consider

I = $\int$ (2 e^x)/(1 + e^(2x)) dx

This integrates to 2arctan(e^x) + C_1.

On the other hand, we can re-write I as:

$\int$ 2/(e^(-x) + e^x) dx

= $\int$ 1/cosh(x) dx

= $\int$ cosh(x)/cosh^2(x) dx

= $\int$ cosh(x)/(1 + sinh^2(x)) dx

= $\int$ 1/(1 + u^2) du ---- (let u = sinh(x))

= arctan(u) + C_2

= arctan(sinh(x)) + C_2

Thus:
arctan(sinh(x)) = 2arctan(e^x) + C --- (where C = C_1 - C_2 is constant)

Now let x=0, and we find that:
0 = 2 * pi/4 + C
=> C = -pi/2

That is,
arctan(sinh(x)) = 2arctan(e^x) - pi/2

I think I'm gonna go take a shower now.

Reply 2

khaixiang

OP

5

I must applaud you for that insight... this is exactly how my problem stemmed from! And I was wondering after I had done the integration, whether one can manipulate $\arctan(\sinh(x))$ to get $2\arctan(e^x)-\frac{\pi}{2}$

Reply 3

DFranklin

18

Maybe I'm missing something, but I think this is not so bad.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$ .

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$ .

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$ .

Reply 4

khaixiang

OP

5

DFranklin

Maybe I'm missing something, but I think this is not so bad.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$ .

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$ .

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$ .

Thanks a bunch... you're right, it's not so bad, in fact it's a bit nice. I was thinking along this line, but God knows why I only thought of tan(A+B) and not tan(A-B) :redface:

Reply 5

dvs

10

It's still nice to see that calculus can be used to prove identities. For another example, if we wanted to prove that sin^2(x) + cos^2(x) = 1, we could set the LHS to f(x) and note that f'(x)=0, i.e. f(x) is constant; then by noting that f(0)=1, we can conclude that f(x)=1. :smile:

Transformation of arctan(sinh(x)) ??

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