These two are giving me a headache...

[Daniel_Feldman]

1. Integrate dx/root(4+root(x))

2. Verify:

(coshx-sinhx)^7=cosh(7x)-sinh(7x) (hyperbolic functions)


On problem one, I thought about rationalizing the denominator and then using the integration of arctan, but that would get a bit messy.

On problem two, I though about a binomial expansion, but that would get quite messy as well.

Any ideas?

 

[pka]

#1 MathCad found NO SOLUTION.
Can you use series?

 

[stapel]

1) This one must be possible (I don't say "easy"), because The Integrator came up with the following:

. . . . .\(\displaystyle \Large{\frac{4}{3}\mbox{ }\left(-8\mbox{ }+\mbox{ }\sqrt{x}\right)\mbox{ }\sqrt{4\mbox{ }+\mbox{ }\sqrt{x}}}\)

Yeah, I was gonna guess that.... :shock:

Eliz.

 

[Daniel_Feldman]

Well, what a beautiful solution. But, alas, the process to get it......drives me insane.


By the way, I had no idea there was an integrator. How cool.

 

[galactus]

1. Integrate dx/root(4+root(x))

Let's try it this way. Hope this works.

$\displaystyle \int(\frac{1}{\sqrt{4+sqrt{x}}})dx$

$\displaystyle u^{2}=4+x^{\frac{1}{2}}$

$\displaystyle 2udu=\frac{1}{2sqrt{x}}dx, u^{2}-4=sqrt{x}$

$\displaystyle 4udu=\frac{1}{sqrt{x}}dx$

$\displaystyle (4u^{3}-16u)du=dx$

$\displaystyle \frac{(4u^{3}-16u)}{u}=(4u^{2}-16)$

Now Integrate:

$\displaystyle \int(4u^{2}-16)du$

=$\displaystyle \frac{4}{3}u^{3}-16u$

=$\displaystyle (\frac{4}{3})(4+\sqrt{x})^{\frac{3}{2}}-16\sqrt{4+\sqrt{x}}$

Whatcha think?. Although it may not appear to, I believe this answer is equivalent to what Stapel posted.

May as well finish simplifying into the other form.

$\displaystyle \frac{4}{3}((4+\sqrt{x})^{\frac{3}{2}})-16\sqrt(4+sqrt{x})$

$\displaystyle \frac{4}{3}(4+\sqrt{x})(\sqrt{4+\sqrt{x}})-16\sqrt{4+\sqrt{x}}$

$\displaystyle (\frac{16}{3}+\frac{4}{3}\sqrt{x})\sqrt{4+\sqrt{x}}-16\sqrt{4+\sqrt{x}}$

$\displaystyle \frac{16}{3}\sqrt{4+\sqrt{x}}+\frac{4}{3}\sqrt{x}\sqrt{4+\sqrt{x}}-16\sqrt{4+\sqrt{x}}$

$\displaystyle \frac{4}{3}\sqrt{x}\sqrt{4+\sqrt{x}}-\frac{32}{3}\sqrt{4+\sqrt{x}}$

$\displaystyle \frac{4}{3}(\sqrt{x}\sqrt{4+\sqrt{x}}-8\sqrt{4+\sqrt{x}})$

\(\displaystyle \frac{4}{3}(\sqrt{x}-8)(\sqrt{4+\sqrt{x})\)

2. Verify:

(coshx-sinhx)^7=cosh(7x)-sinh(7x) (hyperbolic functions)

Are you allowed to use the hyperbolic identities:

$\displaystyle \frac{e^{x}+e^{-x}}{2}=cosh(x)$ and $\displaystyle \frac{e^{x}-e^{-x}}{2}=sinh(x)$

.


On problem one, I thought about rationalizing the denominator and then using the integration of arctan, but that would get a bit messy.

On problem two, I though about a binomial expansion, but that would get quite messy as well.

Any ideas?

 

[Daniel_Feldman]

We can use hyperbolic identities....but I don't see how that would simplify things.

 

[Daniel_Feldman]

Thanks galactus. That looks right, though I keep looking over my work trying to see what I did wrong. I came out to -4u^-2-16 somehow, which gave me a natural log, among other fun things. Meh. Thanks for your help.

 

[soroban]

Hello, Daniel!

2. Verify: .$\displaystyle (\cosh x\,-\,\sinh x)^7\:=\:\cosh(7x)\,-\,\sinh(7x)$

Using the definitions:
. . \(\displaystyle \L\left(\frac{e^x\,+\,e^{-x}}{2}\,-\,\frac{e^x\,-\,e^{-x}}{2}\right)^7\;=\;\left(\frac{2e^{-x}}{2}\right)^7\;=\;\left(e^{-x}\right)^7\;=\;e^{-7x}\)


Then we have: .\(\displaystyle \L e^{-7x}\;=\;\frac{e^{-7x}\,+\,e^{-7x}}{2}\)


In the numerator, add and subtract $\displaystyle e^{7x}:$

. . \(\displaystyle \L\frac{e^{7x}\,+\,e^{-7x}\,-\,e^{7x}\,+\,e^{-7x}}{2}\;=\;\frac{e^{7x}+e^{-7x}}{2}\,-\,\frac{e^{7x}-e^{-7x}}{2}\;=\;\cosh(7x)\,-\,\sinh(7x)\)

 

[soroban]

Hello again, Daniel!

\(\displaystyle \L1.\;\;\int \frac{dx}{\sqrt{4 + \sqrt{x}}}\)

Let $\displaystyle u\,=\,4\,+\,\sqrt{x}\;\;\Rightarrow\;\;x\,=\,(u-4)^2\;\;\Rightarrow\;\;dx\,=\,2(u-4)\,du$

Substitute: .\(\displaystyle \L\int\frac{2(u-4)\,du}{u^{\frac{1}{2}}}\;=\;2\int(u^{\frac{1}{2}}\,-\,4u^{-\frac{1}{2}})\,du\) . . . etc.

 

[Daniel_Feldman]

Awesome. Thanks so much.