Trigonometric Substitution

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thread starter vanilla Apr 10 2009 + PM | QUOTE | PERMALINK | REPORT

A calculus question involving Trigonometric substitution, u substitution, and Power-Reducing Formulas. This stuff is mind boggling..! I hope I didn't make any errors. link

Samomo 04/10/09 + PM | QUOTE | PERMALINK | REPORT

Wut I though calculus would be easy. You did like a 8 step equation.. wut.

sonnyXway 04/10/09 + PM | QUOTE | PERMALINK | REPORT

AffectedWonton said: "1+1= a window im such a genious rite?" And the blob wins the thread!

vanilla Apr 10 2009 + PM | QUOTE | PERMALINK | REPORT

xIridesc3nce said: "So wait... where did the integral of sin^2 w dw go off to? You lost me about halfway through your work when you seemingly just set it aside. If I have my integrals recalled correctly, integral sin^2(x) dx = x/2 + sin(2x)/4 + c. Hopefully that helps you a little bit." Oh, we were proving that the integral turns out that way by using a power reducing formula. By the Power-Reducing Formula, sin�w = (1 - cos 2w)/2 Now, we can replace the function sin�w. Because we can separate the integral, we separate the integral into two parts. We pull the coefficient (1/2) out and we can integrate 1dw which turns out to be w. Mind you, we have coefficients outside the integral. We also pull the coefficient (1/2) out of (cos 2w)/2 At this moment, we do a process called u-substitution. We allow u to equal 2w and du would be 2dw. To properly substitute, we need to have a coefficient of 2 in the integral, we multiply outside the integral (1/2) in order to bring in a coefficient of 2 inside the integral. Now we have (1/2)(1/2) the integral of (cos 2w)(2dw) which now is (1/4) the integral of (cos u du) Integrating cos u du. would result in, sin u. We replace u for its w counterpart so now we have the entire equation to be.. (1/2)w + (1/4)sin 2w + (some constant) We must keep in mind that we had other coefficients outside the integral. So we actually have.. (36/250)w - (36/400)sin 2w + (some constant) I apologize for being a little sloppy on the paper. I was showing a friend how I solved the integral.

xIridesc3nce 04/10/09 + PM | QUOTE | PERMALINK | REPORT

I see. Eh, it's been a while since I've taken math. Nearing a year now, and I guess I've gotten a little rusty with it. Exchanging integrals for cross-aldol products... I personally find it to be a decent exchange for my capacities, though it might not be the same for others.

LeetPhoniex 04/10/09 + PM | QUOTE | PERMALINK | REPORT

Best Handwriting ever

vanilla 04/10/09 + PM | QUOTE | PERMALINK | REPORT

xIridesc3nce said: "I see. Eh, it's been a while since I've taken math. Nearing a year now, and I guess I've gotten a little rusty with it. Exchanging integrals for cross-aldol products... I personally find it to be a decent exchange for my capacities, though it might not be the same for others." Does it seem correct to you so far? I hope so. Because we have a 2w inside the sine, we must use a double-angle formula. (sin 2w) = 2(sin w)(cos w) Because sin w = (opposite)/(hypotenuse) ..and sin w = (5x/6) Therefore, cos w = ((36 - 25x�)^(1/2))/(6) What does w = ? The arcsine of (5x/6) of course. Plug back in all x's that were substituted by w. Cancel things out and simplify your answer.. ..and we get.. (18/125)*(arcsin (5x/6)) - (x)*((36-25x�)^(1/2))*(1/40) + C ..or at least, that's what I think it's supposed to be.. LOL.

KingDragon 04/10/09 + PM | QUOTE | PERMALINK | REPORT

Actually you can just seperate it instead of using trigonometry....................like multiple and divide by 25 and add and subtract 36 to make same as denominator and going further.If you're interested in that method P.M me.

iluvppl 04/10/09 + PM | QUOTE | PERMALINK | REPORT

xkonohamarux 04/10/09 + PM | QUOTE | PERMALINK | REPORT

haha gotta love your trig funtions. i hate proofs.

iusedonuts Apr 10 2009 + PM | QUOTE | PERMALINK | REPORT

Still need help on it?...or did you not need help to begin with 0.o? -too lazy to read work-

xkonohamarux 04/10/09 + PM | QUOTE | PERMALINK | REPORT

btw for those questioning things, there is no 1 way to do a trigonometric proof.

ryannosaur 04/10/09 + PM | QUOTE | PERMALINK | REPORT

I hated this part the most of Calculus...and right when you got to 36/125 int(sin^2 w) you could have just used the property of (1-2cosw)/2 then integrated that in one step to get 36/125 (1/2 w - 1/4 sin (2w) + C) and skipped like 3 or four steps

xlrAcer5 Apr 10 2009 + PM | QUOTE | PERMALINK | REPORT

You a Junior or Sophomore? *btw, I've seen you in maple before :D

AznRyaku Apr 10 2009 + PM | QUOTE | PERMALINK | REPORT

Sovereignity said: "Oh jeez, and I just learned about cosine, sine, and tangent this afternoon. All I know so far is: Tangent = opposite divided by adjacent Sine = Opposite divided by hypotenuse Cosine = Adjacent divided by hypotenuse :/" SoH CaH ToA is a easy way to remember. It is used most times for finding ratios that can be placed into simplest radical form. Confuses me, but its easy. @Jing. All I can say is google it. Or link <-- Goodhelp.

LimusocoBobo 04/10/09 + PM | QUOTE | PERMALINK | REPORT

Pfft... Wait until Calc 3 and Differential Equations ... And don't even get me started on Dynamic Meteorology... @_@