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dinamix

  • one year ago

another challenge who find this integral without use (Binomial theorem ) and google ;p

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  1. dinamix
    • one year ago
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    \[\int\limits_{0}^{\frac{ \pi }{ 2 }} \cos^6x\]

  2. anonymous
    • one year ago
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    You reduce the order several times using the half-angle identity: \[\cos^2x=\frac{1+\cos2x}{2}\] So you have \[\begin{align*} \cos^6x&=\left(\frac{1+\cos2x}{2}\right)^3\\[2ex] &=\frac{1}{8}+\frac{3}{8}\cos2x+\frac{3}{8}\cos^22x+\frac{1}{8}\cos^32x\end{align*}\] and so on.

  3. anonymous
    • one year ago
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    For odd powers of cosine, you can rewrite via the Pythagorean identity: \[\cos^{2k+1}x=\cos x\cos^{2k}x=\cos x(1-\sin^2x)^k\] Change of variables will reduce this sort of expression nicely.

  4. anonymous
    • one year ago
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    Ah but you said without the binomial theorem... Okay. Recall that \[\int_a^b f(x)\,dx=\int_a^bf(a+b-x)\,dx\] This gives \[\int_0^{\pi/2}\cos^6x\,dx=\int_0^{\pi/2}\cos^6\left(\frac{\pi}{2}-x\right)\,dx=\int_0^{\pi/2}\sin^6x\,dx\]

  5. dinamix
    • one year ago
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    problem with 1/8 *(cos^3(2x)) we modif it to 2/16*(cos^3x) i think , i hope understand it

  6. dinamix
    • one year ago
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    sorry its 2/16*(cos^3(2x))*

  7. anonymous
    • one year ago
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    Adding the integral of \(\cos^6x\) to both sides gives \[2\int_0^{\pi/2}\cos^6x\,dx=\int_0^{\pi/2}(\cos^6x+\sin^6x)\,dx\] The RHS contains a sum of cubes: \[\begin{align*} \cos^6x+\sin^6x&=a^6+b^6\\[2ex] &=(a^2+b^2)(a^4-a^2b^2+b^4)\\[2ex] &=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\\[2ex] &=\cos^4x-\cos^2x\sin^2x+\sin^4x \end{align*}\] How you deal with these terms would probably depend on what you mean by "not being able to use the binomial theorem".

  8. dinamix
    • one year ago
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    suppose t=sin2x i think cuz dt= 2cos2x dx i think first answer is easy

  9. dinamix
    • one year ago
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    cuz i see binomial theorem is very long

  10. anonymous
    • one year ago
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    Right, your method for the cubed term is good. \[\int\left(\color{red}{\frac{1}{8}}+\color{red}{\frac{3}{8}\cos2x}+\frac{3}{8}\cos^22x+\color{red}{\frac{1}{8}\cos^32x}\right)\,dx\] The red terms are easy. However, I implicitly used the binomial theorem in order to obtain this expansion from \(\cos^6x\) in the first place.

  11. anonymous
    • one year ago
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    Actually, all the terms are easy with that trig identity...

  12. anonymous
    • one year ago
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    The way I understand your question is that you want to compute the integral without having to expand anything of the form \((a+b)^n\) for \(n\ge2\). Am I right?

  13. dinamix
    • one year ago
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    yup

  14. dinamix
    • one year ago
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    this what i mean dude

  15. dinamix
    • one year ago
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    i think no away without use binomial theorem

  16. anonymous
    • one year ago
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    Right, so the expansion above doesn't abide by your rules. Continuing where I left off: \[\begin{align*} \cos^6x+\sin^6x&=\cos^4x-\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x-\frac{4}{4}\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x-\left(\frac{1}{2}\sin x\cos x\right)^2+\sin^4x\\[2ex] &=\cos^4x-\left(\frac{1}{4}\sin 2x\right)^2+\sin^4x\\[2ex] &=\cos^4x-\frac{1}{16}\sin^2 2x+\sin^4x \end{align*}\] The middle term can be handled with another half-angle identity, \(\sin^2x=\dfrac{1-\cos2x}{2}\). So we have \[\begin{align*}2\int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}(\cos^6x+\sin^6x)\,dx\\[2ex] &=\int_0^{\pi/2}\left(\cos^4x-\frac{1}{32}(1-\cos4x)+\sin^4x\right)\,dx \end{align*}\] Using the same reasoning as before, you have \[\begin{align*}\int_0^{\pi/2}\cos^4x\,dx&=\int_0^{\pi/2}\sin^4x\,dx\\[2ex] 0&=\int_0^{\pi/2}(\cos^4x-\sin^4x)\,dx\\[2ex] &=\int_0^{\pi/2}(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)\,dx\\[2ex] &=\int_0^{\pi/2}\cos2x\,dx\end{align*}\] leaving you with \[\int_0^{\pi/2}\cos^6x\,dx=\frac{1}{64}\int_0^{\pi/2}(\cos4x-1)\,dx\]

  17. anonymous
    • one year ago
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    No wait, there's a mistake somewhere up there...

  18. anonymous
    • one year ago
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    Still looking for the error, but I hope you see the general idea?

  19. freckles
    • one year ago
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    did you change 4/4 to 1/4?

  20. freckles
    • one year ago
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    \[\begin{align*} \cos^6x+\sin^6x&=\cos^4x-\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x-\color{red}{\frac{4}{4}}\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x-\left(\color{red}{\frac{1}{2}}\sin x\cos x\right)^\color{red}{2}+\sin^4x\\[2ex] &=\cos^4x-\left(\frac{1}{4}\sin 2x\right)^2+\sin^4x\\[2ex] &=\cos^4x-\frac{1}{16}\sin^2 2x+\sin^4x \end{align*}\]

  21. anonymous
    • one year ago
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    There it is!

  22. anonymous
    • one year ago
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    Oh actually a much bigger mistake: the integral of \(\cos^4x+\sin^4x\) is not zero.

  23. anonymous
    • one year ago
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    So in fact, we're back to this stage: \[\begin{align*}2\int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}(\cos^6x+\sin^6x)\,dx\\[2ex] &=\int_0^{\pi/2}\left(\cos^4x-\frac{1}{32}(1-\cos4x)+\sin^4x\right)\,dx\\[2ex] &=\int_0^{\pi/2}\left(2\cos^4x-\frac{1}{32}(1-\cos4x)\right)\,dx\end{align*}\] ...which WA is also telling me is not true. Hmm...

  24. dinamix
    • one year ago
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    its 2/2 not 1/2

  25. anonymous
    • one year ago
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    Alright, backing up a bit (again): \[\begin{align*} 2\int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}\left(\cos^4x+\sin^4x-\cos^2x\sin^2x\right)\,dx&(1)\\[2ex] &=\int_0^{\pi/2}\left(\cos^4x+\sin^4x-\frac{1}{4}\sin^22x\right)\,dx&(2)\\[2ex] &=\int_0^{\pi/2}\left(2\cos^4x-\frac{1}{4}\sin^22x\right)\,dx \end{align*}\] (1) http://www.wolframalpha.com/input/?i=2+integral+cos%5E6x+over+0%2Cpi%2F2%3Dintegral+%28sin%5E4x%2Bcos%5E4x-cos%5E2x+sin%5E2x%29+over+0%2Cpi%2F2 (2) http://www.wolframalpha.com/input/?i=cos%5E2x+sin%5E2x (I guess I didn't make a mistake here after all @freckles :P)

  26. dinamix
    • one year ago
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    @freckles

  27. dinamix
    • one year ago
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    is right

  28. dinamix
    • one year ago
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    4/4 will be 2^2/(2^2) and when 2/2 not 1/2 look good

  29. anonymous
    • one year ago
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    \[\begin{align*} \int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}\cos^4x\,dx-\frac{1}{8}\int_0^{\pi/2}\sin^22x\,dx\\[2ex] &=\int_0^{\pi/2}\cos^4x\,dx-\underbrace{\frac{1}{16}\int_0^{\pi/2}(1-\cos4x)\,dx}_{\text{easy}} \end{align*}\]

  30. freckles
    • one year ago
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    A way by integration by parts: \[ \\ \int\limits \cos^6(x) dx= \int\limits \cos(x) \cos^5(x) dx \\ \int\limits \cos(x) \cos^5(x) dx=\sin(x) \cos^5(x)+5 \int\limits \sin^2(x) \cos^4(x) dx \\ \int\limits \cos^6(x) dx=\sin(x) \cos^5(x)+5 \int\limits (1-\cos^2(x)) \cos^4(x) dx \\ \int\limits \cos^6(x) dx=\sin(x) \cos^5(x)+5 \int\limits \cos^4(x) dx -5 \int\limits \cos^6( x) dx \\ 6 \int\limits \cos^6(x) dx= \sin(x) \cos^5(x)+ \int\limits 5 \cos^4(x) dx\] --- \[\int\limits \cos^4(x) dx=\int\limits \cos(x) \cos^3(x) dx \\ =\sin(x) \cos^3(x) + 3 \int\limits \sin^2(x) \cos^2(x) dx \\ =\sin(x) \cos^3(x)+3 \int\limits (1-\cos^2(x)) \cos^2(x) dx \\ =\sin(x) \cos^3(x) +\int\limits 3 \cos^2(x) dx -3 \int\limits \cos^4(x) dx \\ \text{ so .. } \\ 4 \int\limits \cos^4(x) dx=\sin(x) \cos^3(x)+3 \int\limits \cos^2(x) dx \text{ now we can say } \\ \int\limits \cos^6 (x) dx=\sin(x)\cos^5(x)+\frac{5}{4} \sin(x)\cos^3(x)+\frac{15}{4} \int\limits \cos^2(x) dx\] and we can use that double angle identity for that one thingy let me check me work real quick

  31. freckles
    • one year ago
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    oops forgot to divide by 6

  32. freckles
    • one year ago
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    \[6 \int\limits \cos^6(x) dx=\sin(x) \cos^5(x)+\frac{5}{4} \sin(x) \cos^3(x)+\frac{15}{4} \int\limits \cos^2(x) dx \\ \text{ should be the last line }\] then divide by 6

  33. anonymous
    • one year ago
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    Yeah, nice how the first two terms on the right disappear.

  34. anonymous
    • one year ago
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    If you're feeling particularly masochistic, you can also use the tangent half-angle substitution, but I should figure out how to compute \(\displaystyle\int_0^{\pi/2}\cos^4x\,dx\) first (without expanding, of course)...

  35. freckles
    • one year ago
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    \[\int\limits\limits \cos^6(x) dx= \frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x)+\frac{15}{24} \int\limits\limits \frac{1}{2}(1+\cos(2x)) dx \\ \int\limits\limits \cos^6(x) dx=\frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x) +\frac{15}{48}(x+\frac{1}{2} \sin(2x)) +C \]

  36. freckles
    • one year ago
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    its so cute how many different forms the answer can take when we talk about trig functions

  37. freckles
    • one year ago
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    well for the indefinite integral (I ignored the limits in my answer)

  38. anonymous
    • one year ago
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    Here's one way: \[\begin{align*}\cos^4x&=\cos^2x(1-\sin^2x)\\[2ex]&=\cos^2x-\cos^2x\sin^2x\\[2ex]&=\cos^2x-\frac{1}{4}\sin^22x\\[2ex]&=\frac{1+\cos2x}{2}-\frac{1-\cos4x}{8} \end{align*}\] And now integrating should be easy!

  39. anonymous
    • one year ago
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    So... all this work to say, finally, that \[\begin{align*} \int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}\left(\frac{1+\cos2x}{2}-\frac{1-\cos4x}{8}-\frac{1-\cos4x}{16}\right)\,dx\\[2ex] &=\int_0^{\pi/2}\left(\frac{1+\cos2x}{2}-\frac{3(1-\cos4x)}{16}\right)\,dx\end{align*}\] http://www.wolframalpha.com/input/?i=integrate+%28%281%2Bcos2x%29%2F2-3%2F16%281-cos4x%29%29+over+0%2Cpi%2F2%2C+integrate+cos%5E6x+over+0%2Cpi%2F2 Feels good.

  40. dinamix
    • one year ago
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    @freckles u are amazing ,i know method by parts but i think will be hard when use it here , wow super smart ty

  41. anonymous
    • one year ago
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    As useful as identities are, I'll take binomial expansion over them any day :P @freckles

  42. freckles
    • one year ago
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    @dinamix for really really interesting integration questions you should look at @SithsAndGiggles 's profile he is truly a master of integration

  43. dinamix
    • one year ago
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    @SithsAndGiggles , @freckles ty i learn to much rules , but i want see yours opinion about this challenge

  44. anonymous
    • one year ago
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    Thanks @freckles, but I'm hardly a master - there are far more techniques and details yet for me to learn/discover before I'm at that level :3

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