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

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dinamix
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\frac{ \pi }{ 2 }} \cos^6x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You reduce the order several times using the halfangle 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah but you said without the binomial theorem... Okay. Recall that \[\int_a^b f(x)\,dx=\int_a^bf(a+bx)\,dx\] This gives \[\int_0^{\pi/2}\cos^6x\,dx=\int_0^{\pi/2}\cos^6\left(\frac{\pi}{2}x\right)\,dx=\int_0^{\pi/2}\sin^6x\,dx\]

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0problem with 1/8 *(cos^3(2x)) we modif it to 2/16*(cos^3x) i think , i hope understand it

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0sorry its 2/16*(cos^3(2x))*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Adding 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: \[\begin{align*} \cos^6x+\sin^6x&=a^6+b^6\\[2ex] &=(a^2+b^2)(a^4a^2b^2+b^4)\\[2ex] &=(\cos^2x+\sin^2x)(\cos^4x\cos^2x\sin^2x+\sin^4x)\\[2ex] &=\cos^4x\cos^2x\sin^2x+\sin^4x \end{align*}\] How you deal with these terms would probably depend on what you mean by "not being able to use the binomial theorem".

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0suppose t=sin2x i think cuz dt= 2cos2x dx i think first answer is easy

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0cuz i see binomial theorem is very long

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, all the terms are easy with that trig identity...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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?

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0i think no away without use binomial theorem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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 halfangle 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 \[\int_0^{\pi/2}\cos^6x\,dx=\frac{1}{64}\int_0^{\pi/2}(\cos4x1)\,dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No wait, there's a mistake somewhere up there...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still looking for the error, but I hope you see the general idea?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you change 4/4 to 1/4?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align*} \cos^6x+\sin^6x&=\cos^4x\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x\color{red}{\frac{4}{4}}\cos^2x\sin^2x+\sin^4x\\[2ex] &=\cos^4x\left(\color{red}{\frac{1}{2}}\sin x\cos x\right)^\color{red}{2}+\sin^4x\\[2ex] &=\cos^4x\left(\frac{1}{4}\sin 2x\right)^2+\sin^4x\\[2ex] &=\cos^4x\frac{1}{16}\sin^2 2x+\sin^4x \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh actually a much bigger mistake: the integral of \(\cos^4x+\sin^4x\) is not zero.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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%5E4xcos%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)

dinamix
 one year ago
Best ResponseYou've already chosen the best response.04/4 will be 2^2/(2^2) and when 2/2 not 1/2 look good

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} \int_0^{\pi/2}\cos^6x\,dx&=\int_0^{\pi/2}\cos^4x\,dx\frac{1}{8}\int_0^{\pi/2}\sin^22x\,dx\\[2ex] &=\int_0^{\pi/2}\cos^4x\,dx\underbrace{\frac{1}{16}\int_0^{\pi/2}(1\cos4x)\,dx}_{\text{easy}} \end{align*}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1A 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops forgot to divide by 6

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[6 \int\limits \cos^6(x) dx=\sin(x) \cos^5(x)+\frac{5}{4} \sin(x) \cos^3(x)+\frac{15}{4} \int\limits \cos^2(x) dx \\ \text{ should be the last line }\] then divide by 6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, nice how the first two terms on the right disappear.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you're feeling particularly masochistic, you can also use the tangent halfangle substitution, but I should figure out how to compute \(\displaystyle\int_0^{\pi/2}\cos^4x\,dx\) first (without expanding, of course)...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits\limits \cos^6(x) dx= \frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x)+\frac{15}{24} \int\limits\limits \frac{1}{2}(1+\cos(2x)) dx \\ \int\limits\limits \cos^6(x) dx=\frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x) +\frac{15}{48}(x+\frac{1}{2} \sin(2x)) +C \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1its so cute how many different forms the answer can take when we talk about trig functions

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well for the indefinite integral (I ignored the limits in my answer)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here'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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So... 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%2F23%2F16%281cos4x%29%29+over+0%2Cpi%2F2%2C+integrate+cos%5E6x+over+0%2Cpi%2F2 Feels good.

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0@freckles u are amazing ,i know method by parts but i think will be hard when use it here , wow super smart ty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As useful as identities are, I'll take binomial expansion over them any day :P @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@dinamix for really really interesting integration questions you should look at @SithsAndGiggles 's profile he is truly a master of integration

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles , @freckles ty i learn to much rules , but i want see yours opinion about this challenge

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @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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