Integrating cos^6 (x) dx from 0 to pi/2

[s3a]

Homework Statement

This ( http://www.wolframalpha.com/input/?i=integrate+cos^6+(x)+dx+from+0+to+pi/2 ) is the integral I am trying to evaluate.:
int cos^6 (x) dx from 0 to pi/2

Homework Equations

(1 + cos(2x))/2 = cos^2 (x)
(1 – cos(2x))/2 = sin^2 (x)
sin^2 (x) + cos^2 (x) = 1
Variable substitution

The Attempt at a Solution

My (typed-up) work is attached as MyWork.jpg.

I would greatly appreciate it if someone could tell me where I went wrong in my work!

 

[vanhees71]

That's a pretty complicated calculation. I'd rather use the binomial theorem for
\cos x=\frac{\exp(\mathrm{i} x)+\exp(-\mathrm{i} x)}{2}
to evaluate \cos^6 x and then integrate the simple constant and exponential functions!

 

[arildno]

First off:
ALL the antiderivatives that becomes sine will disappear at the limits given (they wil yield integer multiples as arguments within the sine functions.

Thus, the total contributions to the integral is only given by integration of the CONSTANTS, and those, you have not provided.

 

[CAF123]

You missed a term $\int dx$ in the second row and remember that your final answer should be multiplied by 1/8 since you took this out as a factor in line 1.

 

[gopher_p]

You lost your 1/8 from the first line and the integral of 1 from the second line.

Edit: Damn. Sniped.

 

[iRaid]

Isn't it easier to do this:
\int cos^{6}x dx=\int (cos^{2}x)(1-sin^{2}x)(1-sin^{2}x)dx
Then set u=sin²x

 

[arildno]

Not really, iRaid, since we then have:
du=2sin(x)cos(x)dx
This does not neatly eliminate the remaining cos^2(x) factor.

 

[Mandelbroth]

That's a pretty complicated calculation. I'd rather use the binomial theorem for
\cos x=\frac{\exp(\mathrm{i} x)+\exp(-\mathrm{i} x)}{2}
to evaluate \cos^6 x and then integrate the simple constant and exponential functions!

I love complex exponentiation substitutions as much as the next guy, but I think I see an easier substitution.

Remember that $(\cos x)^2=\frac{1+\cos(2x)}{2}$ and $(\sin x)^2=\frac{1-\cos(2x)}{2}$. A bit of algebra leads us to a fairly simple integral, methinks.

 

[s3a]

Edit: Sorry, I double-posted.

 

[s3a]

Vanhees71, I would rather use real numbers (plus, I am expected to)! :D

Also, I took the input the rest of you guys gave and introduced the addition of the integral dx term, in addition to fixing the 1/8 multiplication however, I now get 19π/112 ≠ 5π/32 ... does anyone see what I did wrong now?

 

[Mandelbroth]

... does anyone see what I did wrong now?

Not if you don't show us your work!

 

[s3a]

Not if you don't show us your work!

Lol :D, I realized shortly after my post. Here is the attachment!

 

[arildno]

16*2=28??

 

[D H]

Vanhees71, I would rather use real numbers (plus, I am expected to)! :D

Also, I took the input the rest of you guys gave and introduced the addition of the integral dx term, in addition to fixing the 1/8 multiplication however, I now get 19π/112 ≠ 5π/32 ... does anyone see what I did wrong now?

Hint: Try using Mandelbroth's identities to expand $cos^6(x)$ as $a_0 + a_1\cos(2x) + a_2\cos(4x) + a_6\cos(6x)$. This is trivial with vanhees71's identity. It is a whole lot more work with Mandelbroth's identities, but it possible.

 

[s3a]

Thanks, arildno, that was my last mistake and, I, now, get it perfectly (doing it the way I wanted to do it)!

D H, I don't feel like exploring that too deeply because, I am satisfied with the answer I got but, why would I need to replace the cosines with e bases raised to complex numbers when integrating the cosines from the identity you just gave will simply give me sines that I know will all cancel to 0 and, also, without getting too deep into this and assuming I am correct in my upcoming assumption about $a_0$, how can I find that $a_0 = \frac 5 {16}$ (such that $\frac 5 {16} \frac{\pi}2 = 5 \frac{\pi}{32}$)?

 

[Saitama]

Here's an alternative approach.
Let
$$I=\int_0^{\pi/2} \cos^6x \, dx$$
This is equivalent to
$$I=\int_0^{\pi/2} \sin^6x \, dx$$
Add the two to get
$$2I=\int_0^{\pi/2} ((\cos^2x)^3+(\sin^2x)^3) \, dx$$
Use the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $\sin^2x+\cos^2x=1$ to get
$$2I=\int_0^{\pi/2} (\sin^4x+\cos^4x-\sin^2x \cos^2x)dx$$
Now rewrite $\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$ to get
$$2I=\int_0^{\pi/2} (1-3\sin^2x\cos^2x) \, dx$$
Rewrite $\sin^2x\cos^2x=(1/4)\sin^2(2x)$. Now its a very easy integral to solve.

 

[Mandelbroth]

Here's an alternative approach.
Let
$$I=\int_0^{\pi/2} \cos^6x \, dx$$
This is equivalent to
$$I=\int_0^{\pi/2} \sin^6x \, dx$$
Add the two to get
$$2I=\int_0^{\pi/2} ((\cos^2x)^3+(\sin^2x)^3) \, dx$$
Use the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $\sin^2x+\cos^2x=1$ to get
$$2I=\int_0^{\pi/2} (\sin^4x+\cos^4x-\sin^2x \cos^2x)dx$$
Now rewrite $\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$ to get
$$2I=\int_0^{\pi/2} (1-3\sin^2x\cos^2x) \, dx$$
Rewrite $\sin^2x\cos^2x=(1/4)\sin^2(2x)$. Now its a very easy integral to solve.

That's fantastic! I wouldn't have thought of doing that!

Hint: Try using Mandelbroth's identities...

I get my own identities? I feel like Gauss!

 

[Saitama]

That's fantastic! I wouldn't have thought of doing that!

Thanks!

But honestly, that isn't really fantastic. Its a very standard approach to such kind of problems. :)

 

[D H]

D H, I don't feel like exploring that too deeply because, I am satisfied with the answer I got but, why would I need to replace the cosines with e bases raised to complex numbers when integrating the cosines from the identity you just gave will simply give me sines that I know will all cancel to 0 and, also, without getting too deep into this and assuming I am correct in my upcoming assumption about $a_0$, how can I find that $a_0 = \frac 5 {16}$ (such that $\frac 5 {16} \frac{\pi}2 = 5 \frac{\pi}{32}$)?

The reason is simple. It's rather tedious and extremely easy to make a mistake when using Mandelbroth's identities. When you use vanhees71's identity it is downright trivial. Try integrating cos(x)^1000 from 0 to pi/2 using Mandelbroth's identities. It's *easy* using vanhees71's identity. Read off the constant term as a binomial coefficient and scale appropriately: The answer is ${1000 \choose 500} \frac{\pi}{2^{1001}}$.

 

[sankalpmittal]

The reason is simple. It's rather tedious and extremely easy to make a mistake when using Mandelbroth's identities. When you use vanhees71's identity it is downright trivial. Try integrating cos(x)^1000 from 0 to pi/2 using Mandelbroth's identities. It's *easy* using vanhees71's identity. Read off the constant term as a binomial coefficient and scale appropriately: The answer is ${1000 \choose 500} \frac{\pi}{2^{1001}}$.

I have two approach:

1) cos⁶x=(cos³x)² = (cos3x+3cosx)²/16 = ...

Keep applying formulas of cos²θ = (1+cos2θ)/2 and 2cosAcosB= cos(A+B)+cos(A-B) and you'll get easily integrable integrand.

2) Make use of the property of definite integration.

If F(x) is the integrand then F(x)=F(a+b-x) where a and b are upper and lower limit of integration.
Now add to get two times integral I and solve for it.