Mega hard integral <plz help>

[caseclosed]

any suggestions? I think have to use partial fraction decomposition but I don't seem to get anywhere...

[jordan]

Let x equal tan^6.

[insane_alien]

tan^6 means nothing. tan^6 of what ?

[caseclosed]

Let x equal tan^6.

thank you, I was letting x^(1/6)=tan(x) so the problem got really complicated and I ended up nowhere, lol....

 

now here is another one I have trouble on, no clue where to start this one at all

integrate 1/(1+cos(x)+sin(x))

[Tartaglia]

Good spot by Jordan

 

With x = tan^6(A) the integral simplifies to

 

/

|6tan^5(A)sec^2(A) dA/((tan^3(A)*(1+tan^2(A)))=

/

 

/

|6tan^2(A)dA

/

 

1/(1+cos(x)+sin(x)) can be done by substituing sinx = 2t/1+t^2 and Cosx = (1-t^2)/(1+t^2) where t = tan(0.5x)

[Klaynos]

I'd suggest you are seriousely over complicating this.

 

if you just times out the root(x) =>

 

[math] \int (x^\frac {1}{2} + x^\frac {5}{6})^{-1} dx[/math]

 

Which is easier... I think :s

 

although I might be talking rubbish :s

[the tree]

Assuming Klay is right, [math]\int(x^{\frac{1}{2}}+x^{\frac{5}{6}})^{-1}\cdot dx=\int x^{\frac{-1}{2}}+x^{\frac{-5}{6}}\cdot dx= \frac{x^\frac{1}{2}}{\frac{1}{2}}+\frac{x^\frac{1}{6}}{\frac{1}{6}}+c[/math] I think.

[Klaynos]

I'd say there's a chance you've done the final integral wrong, not taken the -ve's into account...

[Tartaglia]

Since the answer is

 

6x^(1/6) + 6*arctan(x^(1/6)) + C

 

I would say a tan substitution is necessary

[NeonBlack]

tree- that's not quite how exponents work.

[the tree]

meh, care to elaborate?

[NeonBlack]

If I used what you did for exponents, [math](a+b)^2 = a^2+b^2[/math]

You can't distribute an exponent. If you have [math](a+b)^{-1}[/math] You cannot say that this is [math]a^{-1}+b^{-1}[/math]

That would be the same as saying [math]\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}[/math]

[the tree]

eep, my bad. I knew that, just missed it.

[Tom Mattson]

I'd suggest you are seriousely over complicating this.

 

if you just times out the root(x) =>

 

[math] \int (x^\frac {1}{2} + x^\frac {5}{6})^{-1} dx[/math]

 

Which is easier... I think :s

 

although I might be talking rubbish :s

 

I do agree that this has been overcomplicated' date=' but I don't see how your suggestion helps.

 

I would say a tan substitution is necessary

 

It isn't. All you have to do is let [imath]u=\sqrt[6]{x}[/imath] in the original integral. Then you can do a partial fraction expansion, as caseclosed had initially surmised.

 

That solution seems a whole lot more obvious to me, but to each his own...

[Klaynos]

I do agree that this has been overcomplicated' date=' but I don't see how your suggestion helps.

 

 

 

It isn't. All you have to do is let [imath']u=\sqrt{x}[/imath] in the original integral. Then you can do a partial fraction expansion, as caseclosed had initially surmised.

 

That solution seems a whole lot more obvious to me, but to each his own...

 

 

I agree I wasn't thinking when I posted it seems. Sorry for any confussion I caused...

[Tartaglia]

Tom - My mind doesn't actually work that way. I just thought Jordan's original idea was smart and unusual

 

The point about the tan substitution is that regardless as to how you do it the final arctan term comes from a direct or implied tan substitution. I didn't really mean to suggest the tan^6(A) substitution was the only one possible

[Tom Mattson]

All you have to do is let [imath]u=\sqrt{x}[/imath] in the original integral. Then you can do a partial fraction expansion' date=' as caseclosed had initially surmised.

[/quote']

 

FYI, there's a typo there. I meant for you to let u equal the 6th root of x, not the square root.

[Dave]

Edited that for you so that people don't get confused later on