1. ## Hyperbolic Trig Proof

Hey all,

I'm supposed to show that Sec (Theta) = Cosh (Theta)

Given:

x = ln (Sec (theta) + tan (theta))

I have NO idea where to start, any pointers?

-B

P.S.

I'm supposed to compute the integral of x^5(Sin(x^3)). I should use parts to integrate this, correct?

2. But Sec(T) is not equal to Cosh(T), for instance:

Sec(Pi) = -1,
Cosh(Pi) = 11.59

Are you sure you have written down the problem correctly?

3. With regards to your P.S.,

use the substitution:

u = x^3,

then you need only integrate by parts once.

4. Originally Posted by TyrsFromAbove37
P.S.

I'm supposed to compute the integral of $x^5sin(x^3)$. I should use parts to integrate this, correct?
Nope, at least not right away. Do a u-substitution, letting $t=x^3$. Thus $dt=3x^2\,dx$

Now $\int x^5sin(x^3)\,dx = \frac{1}{3}\int x^3sin(x^3)*3x^2\,dx = \frac{1}{3}\int tsin(t)\,dt.$.

NOW you use integration by parts (letting $u=t$ and $dv = sin(t)$)to obtain $\frac{1}{3}\int tsin(t)\,dt = \frac{1}{3}(sin(t)-tcos(t))$.

Plugging back in for t yields: $\frac{1}{3}sin(x^3)-\frac{1}{3}x^3cos(x^3)$.

Thus: $\int x^5sin(x^3)\,dx = \frac{1}{3}sin(x^3)-\frac{1}{3}x^3cos(x^3) + C$

5. Originally Posted by Mentia
But Sec(T) is not equal to Cosh(T), for instance:

Sec(Pi) = -1,
Cosh(Pi) = 11.59
I thought it had something to do with the ln function, possibly dealing with the derivatives of the trigonometric functions as they related to a fraction, which would eliminate the trig function in and of itself and leave the fraction with only constants and x terms...(Hope that made sense)

At any rate, I'll ask the prof tomorrow to clarify, maybe he miswrote the problem...

Thanks,

-B

6. Regarding your actual question, I believe I have the solution.

Take the derivative of both sides of your equation:

$\frac{d}{dx}[x] = \frac{d}{d\theta}[ln(sec(\theta)+tan(\theta))]$

Thus, $1 = sec(\theta)$. (Yeah, that complicated expression on the right is just $\int sec(\theta)\,d\theta$.)

Since $1=sec(\theta)$, $\theta = arcsec(1) = 0$.

Thus $sec(0) = cosh(0) = 1$.

QED

7. Thanks for all the help guys, it is greatly appreciated...

-B

8. Did you simply want a value for which Sec(T) = Cosh(t)? There are an infinite number of them, you can show that

$T=ln(4- { e}^{ i+1} - {e }^{i-1 } - {e }^{1-i } - {e }^{1+i } )$

but remember that the complex natural log is a multivalued function. In putting it in this form I may have even gotten rid of some solutions... anyone have some pointers on finding all solutions?

9. I was able to clarify what my prof wanted for the sec = cosh problem:

He wanted us to take the value for cosh (having to do with the e^x terms)

and relate it to the ln function (as it pertains to e^x terms)

Hope that makes sense...

At any rate, I tried plugging the values into the cosh function, but didn't get an answer or I didn't get an answer that made the problem look done...

Any thoughts?

-B

10. Ok i figured it out. You did write it down slightly wrong. What he wants is for you to show Cosh(x) = Sec(T).

You can raise e by both sides of your equation. Then you know e^x. You also know e^-x. Then add them. You can solve it down to Cosh(x) = Sec(T).

11. I'm a bit confused by your wording. Mind clarifying what you mean when you talk about raising both sides by e?

Thanks,

-B

12. For example if we had:

x = ln (y)

Then raise e to the power of both sides:

e^x = e ^ [ln (y)] = y

This is the same as saying:

If 2 = 2 then e^2 = e^2. Right?

Doing this is often used if you have a natural log since it is the inverse natural log:

e^[ln (x)] = x

13. I applied that to the cosh function and ended up getting:

(Sec(theta)+Tan(theta)) - (Sec(theta)+Tan(theta)) / 2

This cancels out to 0...I don't know where to take it from there...

-B

14. You should get:

${e }^{ x} = ( \frac{ 1}{ cos( \theta )} )(1+sin( \theta ))$

Then this means that:

${e }^{ -x} = (cos( \theta )) {(1+sin( \theta )) }^{ -1}$

Then
$
{e }^{ x} + {e }^{ -x} = ( \frac{ 1}{ cos( \theta )} )(1+sin( \theta )) + (cos( \theta )) ( \frac{ 1}{ 1 + sin( \theta )} )$

Now get a common denominator for the right side of the equation and solve it down. You should find that the right side is equal to 2/cos(T). Then divide both sides by 2 and you find Cosh(x) = Sec(T)

15. Alright, that makes sense, thanks for the help.

-B

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