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

1. dinamix

$\int\limits_{0}^{\frac{ \pi }{ 2 }} \cos^6x$

2. anonymous

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

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

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

problem with 1/8 *(cos^3(2x)) we modif it to 2/16*(cos^3x) i think , i hope understand it

6. dinamix

sorry its 2/16*(cos^3(2x))*

7. anonymous

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

suppose t=sin2x i think cuz dt= 2cos2x dx i think first answer is easy

9. dinamix

cuz i see binomial theorem is very long

10. anonymous

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

Actually, all the terms are easy with that trig identity...

12. anonymous

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

yup

14. dinamix

this what i mean dude

15. dinamix

i think no away without use binomial theorem

16. anonymous

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

No wait, there's a mistake somewhere up there...

18. anonymous

Still looking for the error, but I hope you see the general idea?

19. freckles

did you change 4/4 to 1/4?

20. freckles

\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

There it is!

22. anonymous

Oh actually a much bigger mistake: the integral of $$\cos^4x+\sin^4x$$ is not zero.

23. anonymous

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

its 2/2 not 1/2

25. anonymous

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

@freckles

27. dinamix

is right

28. dinamix

4/4 will be 2^2/(2^2) and when 2/2 not 1/2 look good

29. anonymous

\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

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

oops forgot to divide by 6

32. freckles

$6 \int\limits \cos^6(x) dx=\sin(x) \cos^5(x)+\frac{5}{4} \sin(x) \cos^3(x)+\frac{15}{4} \int\limits \cos^2(x) dx \\ \text{ should be the last line }$ then divide by 6

33. anonymous

Yeah, nice how the first two terms on the right disappear.

34. anonymous

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

$\int\limits\limits \cos^6(x) dx= \frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x)+\frac{15}{24} \int\limits\limits \frac{1}{2}(1+\cos(2x)) dx \\ \int\limits\limits \cos^6(x) dx=\frac{1}{6} \sin(x) \cos^5(x)+\frac{5}{24} \sin(x) \cos^3(x) +\frac{15}{48}(x+\frac{1}{2} \sin(2x)) +C$

36. freckles

its so cute how many different forms the answer can take when we talk about trig functions

37. freckles

well for the indefinite integral (I ignored the limits in my answer)

38. anonymous

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

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

@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

As useful as identities are, I'll take binomial expansion over them any day :P @freckles

42. freckles

@dinamix for really really interesting integration questions you should look at @SithsAndGiggles 's profile he is truly a master of integration

43. dinamix

@SithsAndGiggles , @freckles ty i learn to much rules , but i want see yours opinion about this challenge

44. anonymous

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