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Integral of -64cos^3x

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Original post by Zacken
:yes:



Flattering but very untrue. :lol:

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Awww thank you. Just makes me nervous as I'm doing maths at uni but there's people like you around! So what do I know! Oh well


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Reply 21
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Zacken
22
Original post by maths_4_life
Why can't you? Thanks though


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Well, you're integrating with respect to x. And sin⁡x\sin x is a function of xx. Would you agree that ∫x dx=x∫1 dx=x(x+c)=x2+cx\int x \, \mathrm{d}x = x \int 1 \, \mathrm{d}x = x (x +c) = x^2 + cx? Of course not - x2x^2 is a function of xx (that is, it depends and varies as xx varies) and can't be treated as a constant. Same applies to sin⁡x\sin x.
Original post by Student403
That's the key though. You asked and that makes you a really good learner. So keep using TSR (maths forum - stay away from chat :afraid:) and you'll be fine! :awesome:


:smile:))


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Original post by Zacken
Well, you're integrating with respect to x. And sin⁡x\sin x is a function of xx. Would you agree that ∫x dx=x∫1 dx=x(x+c)=x2+cx\int x \, \mathrm{d}x = x \int 1 \, \mathrm{d}x = x (x +c) = x^2 + cx? Of course not - x2x^2 is a function of xx (that is, it depends and varies as xx varies) and can't be treated as a constant. Same applies to sin⁡x\sin x.


I don't really understand your example :/ can you explain a bit more....like how it is similar to the problem I had


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Original post by Zacken
Well, you're integrating with respect to x. And sin⁡x\sin x is a function of xx. Would you agree that ∫x dx=x∫1 dx=x(x+c)=x2+cx\int x \, \mathrm{d}x = x \int 1 \, \mathrm{d}x = x (x +c) = x^2 + cx? Of course not - x2x^2 is a function of xx (that is, it depends and varies as xx varies) and can't be treated as a constant. Same applies to sin⁡x\sin x.


Ohhhh no I understand!!!! Just overcomplicating things! Thank you that really helps


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Zacken
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Original post by maths_4_life
I don't really understand your example :/ can you explain a bit more....like how it is similar to the problem I had


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You asked "why can't you pull sin⁡x\sin x out of the integral", right? The answer is because it's a function of xx and you're integrating with respect to xx.
Original post by maths_4_life
Why can't you? Thanks though
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Because of what dxdx means, it's a little complicated to go into now, but notice that clearly:
∫x dx=x22+C≠x∫1 dx=x2+Cx.\displaystyle\int x \ dx = \dfrac{x^2}{2}+\mathcal{C} \not = x \int 1 \ dx =x^2+\mathcal{C}x.

EDIT: Wow, I see Zacken has this covered :/ at least my constant is prettier.
(edited 8 years ago)
Original post by joostan
Because of what dxdx means, it's a little complicated to go into now, but notice that clearly:
∫x dx=x22+C≠x∫1 dx=x2+Cx.\displaystyle\int x \ dx = \dfrac{x^2}{2}+\mathcal{C} \not = x \int 1 \ dx =x^2+\mathcal{C}x.

EDIT: Wow, I see Zacken has this covered :/ at least my constant is prettier.


Haha yes thank you!!


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Reply 28
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Zacken
22
Original post by joostan
Because of what dxdx means, it's a little complicated to go into now, but notice that clearly:


I'm not sure if you've seen this before but I thought it was very interesting (I did get lost about 2 paragraphs in, but it's more suitable for you level).

P.S: I'm glad to see I used the same counterexample as you. :biggrin:
Original post by joostan
at least my constant is prettier.


Rekt
Original post by Student403
Rekt

Hehe

Original post by Zacken
I'm not sure if you've seen this before but I thought it was very interesting (I did get lost about 2 paragraphs in, but it's more suitable for you level).

P.S: I'm glad to see I used the same counterexample as you. :biggrin:


I saw you post it before, there's ups and downs for me, didn't take Geometry so manifolds are aren't my thing :colondollar:.

P.S: I'm glad to see I used the same counterexample as you. :biggrin:

Hehe, simplest one there is :tongue:.

A problem for those interested might be to find all functions such that:
x∫f(x) dx=∫xf(x) dxx\displaystyle\int f(x) \ dx =\int xf(x) \ dx.
(edited 8 years ago)
Reply 31
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Zacken
22
Original post by joostan

A problem for those interested might be to find all functions such that:
x∫f(x) dx=∫xf(x) dxx\displaystyle\int f(x) \ dx =\int xf(x) \ dx.


Differentiate both sides: xf(x)+∫f(x) dx=xf(x)  ⟺  ∫f(x) dx=0⇒f(x)=0x f(x) + \int f(x) \, \mathrm{d}x = xf(x) \iff \int f(x) \, \mathrm{d}x = 0 \Rightarrow f(x) = 0? Too late for me, I think, can't spot what I've done wrong. :tongue:
Original post by Zacken
Differentiate both sides: xf(x)+∫f(x) dx=xf(x)  ⟺  ∫f(x) dx=0⇒f(x)=0x f(x) + \int f(x) \, \mathrm{d}x = xf(x) \iff \int f(x) \, \mathrm{d}x = 0 \Rightarrow f(x) = 0? Too late for me, I think, can't spot what I've done wrong. :tongue:


No, you're not wrong, I must've misremembered the question I was going to ask. . . it's a post somewhere on here, though I don't know if I can find it now :/
If you replace xx with g(x)g(x) you can actually show it's only true for constants.
Reply 33
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Zacken
22
Original post by joostan
No, you're not wrong, I must've misremembered the question I was going to ask. . . it's a post somewhere on here, though I don't know if I can find it now :/
If you replace xx with g(x)g(x) you can actually show it's only true for constants.


Aye, I remember being the one asking you about something and you bringing that up (if that's what you're talking about) - it was:

∫(f′(x))2 dx=f(x)2+C\displaystyle \int (f'(x))^2 \, \mathrm{d}x = f(x)^2 + \mathcal{C}

Yeah?
Original post by Zacken
Aye, I remember being the one asking you about something and you bringing that up (if that's what you're talking about) - it was:

∫(f′(x))2 dx=f(x)2+C\displaystyle \int (f'(x))^2 \, \mathrm{d}x = f(x)^2 + \mathcal{C}

Yeah?


Possibly., in fact quite probably . . I'm getting old, forgetting things :frown:
Reply 35
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Zacken
22
Original post by joostan
Possibly., in fact quite probably . . I'm getting old, forgetting things :frown:


#SecondYearMathmoThings :lol:
Original post by Zacken
I know you've been taught that you add one to the power, divide by the new power and the divide by the derivative. But that's just plain incorrect. It only applies to integrands of the form (ax+b)n(ax+b)^n. Anything else, that "rule" doesn't hold anymore and you need to be a whole lot more creative. Differentiation is easy (squeezing toothpaste out of the tube) and integration is hard (putting the toothpaste back into the tube), there are no straightforward rules for it. The above users have given a nice way to integrate this function.


Just an after thought ..... Would the method I had used work for anything in the form (af(X) +b)^n or only (ax+b)^n ?
Eg would it work for (asin(X) + b)^n ? Or (ax^2 + b)^n ?
Thanks


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Original post by maths_4_life
Just an after thought ..... Would the method I had used work for anything in the form (af(X) +b)^n or only (ax+b)^n ?
Eg would it work for (asin(X) + b)^n ? Or (ax^2 + b)^n ?
Thanks


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No, the term must be linear. More generally we require an additional term.
∫f′(x)(af(x)+b)n dx=(af(x)+b)n+1a(n+1)+C\displaystyle \int f'(x)(af(x)+b)^n \ dx = \dfrac{(af(x)+b)^{n+1}}{a(n+1)}+\mathcal{C}.
Reply 38
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Zacken
22
Original post by maths_4_life
Just an after thought ..... Would the method I had used work for anything in the form (af(X) +b)^n or only (ax+b)^n ?
Eg would it work for (asin(X) + b)^n ? Or (ax^2 + b)^n ?
Thanks


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And if you're wondering why Joostan's formula works, the substitution u=f(x)u = f(x) will prove illuminating.
Original post by B_9710
∫cos⁡3xdx=∫cos⁡x(cos⁡2x)dx=∫cos⁡x(1−sin⁡2x)dx=∫(cos⁡x−sin⁡2xcos⁡x)dx \displaystyle \int \cos^3 x dx =\int \cos x (\cos^2 x ) dx = \int \cos x (1-\sin^2 x) dx = \int \left ( \cos x -\sin^2 x\cos x \right ) dx .


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