Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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
\[\int_0^{\pi/2}\cos^6x\,dx=\int_0^{\pi/2}\cos^6\left(\frac{\pi}{2}-x\right)\,dx=\int_0^{\pi/2}\sin^6x\,dx\]

Answer 5

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

Answer 6

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

Answer 7

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".

Answer 8

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

Answer 9

cuz i see binomial theorem is very long

Answer 10

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.

Answer 11

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

Answer 12

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?

Answer 13

yup

Answer 14

this what i mean dude

Answer 15

i think no away without use binomial theorem

Answer 16

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
\[\int_0^{\pi/2}\cos^6x\,dx=\frac{1}{64}\int_0^{\pi/2}(\cos4x-1)\,dx\]

Answer 17

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

Answer 18

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

Answer 19

did you change 4/4 to 1/4?

Answer 20

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

Answer 21

There it is!

Answer 22

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

Answer 23

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...

Answer 24

its 2/2 not 1/2

Answer 25

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)

Answer 26

@freckles

Answer 27

is right

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

oops forgot to divide by 6

Answer 32

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

Answer 33

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

Answer 34

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)...

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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!

Answer 39

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.

Answer 40

@freckles u are amazing ,i know method by parts but i think will be hard when use it here , wow super smart ty

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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