/sci/ - Science & Math

[Anonymous No.12703537]

How do I solve this, without bothering to open the paranthesis? like, is there any substitiution possible to change it? or at least do somthing about the cubic power?

[Anonymous No.12703574]

Expand and deal with each piece as a separate integral.

[Anonymous No.12703577]

>>12703574
Thank you for doing his homework

[Anonymous No.12703583]

>>12703537
Use trig identities bro. sinx = 2sin(x/2) cos(x/2)

[Anonymous No.12703587]

>I dont do peoples homework for them why would you think that?

[Anonymous No.12703603]

>>12703583
I tried to do so to make a ((cosx/2 + sinx/2)^2)^3, but it was a futile attempt and i have no idea what to do next

>>12703574
thats even longer than expanding it in the begining and I really want to avoid that.

[Anonymous No.12703607]

>>12703603
What your asking for isn't something you can really do. It's like asking how you cook a can of baked beans and stipulating that you don't want to open the can up.

[Anonymous No.12703621]

>>12703607
sometimes it's possible to solve this kind of equations using eulers identity and i was just wondering if thats the case here as well.

[Anonymous No.12703628]

>>12703621
*integrations

[Anonymous No.12703646]

>>12703537
The easiest way is to involve branch cuts and enlist the Cauchy principal value of the integral. Once you apply the Schlaefli integral, then you use Stirling’s formula in coordination with the infinite sum representation of the trig functions (taking special care to apply residue theorem when necessary).

[Anonymous No.12703649]

>>12703621
Also you meant to say Euler’s formula. You can do that, but it’ll mean you have to apply the method of stationary phase.

[Anonymous No.12703650]

>>12703603
That's the right track. Now the whole integral is merely (Cos(x/2) + sin(x/2)^(6). Next use the addition identity for sine and cosine.

[Anonymous No.12703657]

>>12703650
This guy is giving you wrong answers. Do this>>12703646 or this
>>12703649
instead

[Anonymous No.12703660]

>>12703587
If you keep posting wojacks I will start replying to every homework thread, out of sheer spite and pettiness. I mean.. do you picture the wojack evoking a specific response from the anons you reply to? What is the intended effect, when you post them? It's alright if you tell me, this is an anonymous forum.

[Anonymous No.12703661]

>>12703660
I post them because they have funny face

[Anonymous No.12703678]

>>12703657
Don't confuse him. The original anon was right. Cauchy is overkill here and unnecessary

[Anonymous No.12703681]

>the best on my calculus class
>6 years since i last saw or did an integral
>Can't for the hell of it remember how this shit is done
Holy crap, gotta read some book to catch up. Any recommendations on books? Hopefully with exercises in it, extra internet points if it has the solutions of the exercises.

[Anonymous No.12703689]

>>12703681
who cares about being able to solve an integral, just reach into your bag of tricks and integrate woweezowie

[Anonymous No.12703708]

>>12703537
you can try [math]\alpha = \tan x[/math]

[Anonymous No.12703728]

2sin(x) + cos(x) = Acos(x + B)

[Anonymous No.12703732]

>>12703646
thanks. it's done but i still dont really understand it. I'll have to study more

>>12703650
unfortunatly, it was not the case. you can't turn the original equation into this.

>>12703657
thanks.

[Anonymous No.12703774]

>>12703728
took way too long for someone to suggest the obvious lmao

[Anonymous No.12703783]

>>12703537
>>12703708
[eqn]\sin(x) = \frac{\alpha}{\pm\sqrt{\alpha^2+1}}[/eqn]
[eqn]\cos(x) = \frac{1}{\pm\sqrt{\alpha^2+1}}[/eqn]
[eqn]dx = (\frac{1}{\alpha^2+1})d\alpha[/eqn]
then we get a bunch of integrals like
[eqn]\pm\int (2 \alpha + 1)^3 (\alpha^2+1)^{-5/2}\, d\alpha = \pm\left(\frac{14 \, \alpha^{3} - 24 \, \alpha^{2} + 3 \, \alpha - 22}{3 \, {\left(\alpha^{2} + 1\right)}^{\frac{3}{2}}}\right) + C[/eqn]
[eqn]\pm\int (2 \alpha - 1)^3 (\alpha^2+1)^{-5/2}\, d\alpha = \pm\left(\frac{-14 \, \alpha^{3} - 24 \, \alpha^{2} - 3 \, \alpha - 22}{3 \, {\left(\alpha^{2} + 1\right)}^{\frac{3}{2}}}\right) + C[/eqn]
stitch them up and ur good to go

[Anonymous No.12703797]

>>12703537
This can be written as a linear combination of the first four Legendre polynomials with argument cos(x). Then the solution, using the orthonormality of the Legendre polynomials, is trivial.

[Anonymous No.12704517]

>>12703537
.gg/TJvGnYfknX