Help me with this math problem.

[My Opinion Matters]

(1/2,0)∫ 5 sin^7 2∏x dx

Show your work; but remember, this board has a 25000 character limit for each post.

Aaaaand go! :texan:

[Dio]

eleventeen

[NormanPride]

http://homepage.eircom.net/~miscellaneous/images/animath.jpg

[yermom]

what is (1/2,0)?

and should it be 5*sin^7(2*pi*x) dx?

[My Opinion Matters]

what is (1/2,0)?

should it be 5*sin^7(2*pi*x) dx?

Those are the limits of integration, you silly man.

[OUMallen]

I wish I were your derivative so that I could lie tangent to your curves.

[yermom]

Those are the limits of integration, you silly man.

.5 to 0?

[My Opinion Matters]

And yes, 5 times sin raised to the 7th power times (2pi*x), integrated. Quit stalling!

[My Opinion Matters]

.5 to 0?

Yes.

[yermom]

i don't know if i care enough to go this far back in time :D

[soonerchk]

3. The answer is always 3.

[My Opinion Matters]

3. The answer is always 3.

I don't think this is right.

[soonerchk]

I don't think this is right.

Then it's 7. Those are the only options.

[Fraggle145]

http://homepage.eircom.net/~miscellaneous/images/animath.jpg

This doesnt work. -7 doesnt equal 19. Unless there are subtle differences in the x's that I'm not catching.

[Thaumaturge]

That's a pretty standard type of integral. Why would you come to these rubes?

[GKeeper316]

the answer is 42. duh.

[Howzit]

3. The answer is always 3.

B got me through Physics 2 in college.

[MR2-Sooner86]

I'll get you started:

∫sin^5(x)dx
∫sin(x)sin^2(x)sin^2(x)dx
∫sin(x) (1-cos^2(x)) (1-cos^2(x)) dx
∫sin(x) (1 - 2cos^2(x) + cos^4(x)) dx
∫(1 - 2(cos(x))^2 + (cos(x))^4) sin(x)dx
Let u = cos(x). Then du = -sin(x)dx. So, we have:
∫(1 - 2u^2 + u^4) * -du
-∫(1 - 2u^2 + u^4)du
-(u - 2(u^3)/3 + (u^5)/5)

I think you can take it from there.

[crawfish]

Can I use imaginary numbers?

[GKeeper316]

42

[My Opinion Matters]

I'll get you started:

∫sin^5(x)dx
∫sin(x)sin^2(x)sin^2(x)dx
∫sin(x) (1-cos^2(x)) (1-cos^2(x)) dx
∫sin(x) (1 - 2cos^2(x) + cos^4(x)) dx
∫(1 - 2(cos(x))^2 + (cos(x))^4) sin(x)dx
Let u = cos(x). Then du = -sin(x)dx. So, we have:
∫(1 - 2u^2 + u^4) * -du
-∫(1 - 2u^2 + u^4)du
-(u - 2(u^3)/3 + (u^5)/5)

I think you can take it from there.

Close, but no.

[Partial Qualifier]

Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):

Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).

So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.

J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).

However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).

However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).

I hope this helps.

[afs]

NEEERRRDDDSSS!!!

http://3.bp.blogspot.com/_I2i8xqihwqQ/TDKwKDZiTJI/AAAAAAAAAXU/PBKhmWc2Dqo/s320/revenge-of-the-nerds.jpg

[Howzit]

Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):

Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).

So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.

J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).

However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).

However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).

I hope this helps.

Beat me to it.

[My Opinion Matters]

Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):

Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).

So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.

J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).

However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).

However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.

Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).

I hope this helps.

I am going to have to say this is probably right...because I am not adequately qualified to say that it is not.

[yermom]

Close, but no.

it was close enough that i remembered how to do it

[tommieharris91]

This thread needs more politics and mundane ****-flinging.

[soonerboomer93]

Wtf is Ike when we need a math guy

[yermom]

This thread needs more politics and mundane ****-flinging.

Leibniz was a little bitch, and couldn't sniff Newton's jock

[yermom]

Wtf is Ike when we need a math guy

meh. this is Calc II. this is below me, i know it's below Ike :D

i've had literally 7 or 8 math courses since doing this...

[Partial Qualifier]

Leibniz was a little bitch, and couldn't sniff Newton's jock

....and Einstein wiped his *** with Principia!1!!one!11!!!! ;)

[badger]

42

Ruuuuuuuuuuuuuuuuuuu!

The answer on defense was always 42 back then :D

[stoopified]

0.I did it all in my head.

[sanantoniosooner]

0.I did it all in my head.

dang if this isn't a softball:D

[StoopTroup]

Anyone have Ginger lee's Digits?

[badger]

http://static.gamesradar.com/images/mb/GamesRadar/us/Games/P/Professor%20Layton%20and%20the%20Unwound%20Future/Bulk%20Viewer/DS/2010-06-15/NTR_PLaytonUF_06ss07_E3--article_image.jpg

[swardboy]

E= Mc2
Next.