Integrals: Trig Substitution 3 (long problem)

[shonaragibson]

OHMAGAD I GET IT NOW! a week in university = 18 minutes on youtube.

[NOMVrewq]

Great video! I just have 2 questions. Should the «2» become a «+ or - 2» when you pull it out from the square root at 3:35? Does it even matter?

[Bakmoon]

@NOMVrewq It stays positive. You only add a plus minus sign when you get rid of the square root sign.

[hmmmmusic]

Thank you for your video! I have a few questions :)

how do you decide to set ((x-3)/2)^2 equal to sin^2 theta?

Can't it also equal to cos^2theta? b/c 1-cos^2theta= sin^2theta and  1-sin^2theta=cos^2theta. and you are simply substituting sin^2theta or cos^2theta in the formula 1-u^2?

Thank you Prof Khan!

[Shashankpal1994]

@hmmmmusic

This will solve ur problem:

(en(dot0wikipedia(dot)org/wiki­/Trigonometric_substitution#Su­bstitutions_that_eliminate_tri­gonometric_functions)

[llx2o1]

i seriously had a headache from following this...

[362wowwow]

nice. i had the same problem. the only difference is that mine is a definite integral. SMH. -_-

[desertfox1792]

@khanacademy what program do you use for your videos to write on?

[sonicrange2]

i m just wondering would u be able to simply just get to ur purple simplification (at 4:20) and then substitute for (x-3)= u and figureout du in terms of dx and just go from there it seems easier than a trig substitution although im not sure if it works i havent tried

[tringuyen552911]

@9:43, it's supposed to be 2theta + 2sin of 2theta + C... You missed the 2nd (2)

[ReaverKS]

@tringuyen552911 Recall that to do d/dtheta (sin2(theta)) you must utilize the chain rule, so first you take the dervitative of the innermost portion, 2theta, which is 2, and then the derivative of the outer part, sin, which is cos, but carry the original inner through, so the derivative of sin2theta is 2cos2theta, which is what you originally have in the integral, so Sal did it right.

[ivanmayorov2]

2theta - sin 2 theta + c, not +!!!! :)

[jmlrugby11]

why didn't he just use the arcsin integral? he made it way more complicated then it really is

[jmlrugby11]

@jmlrugby11 scratch that you need 1/a^2 + u^2

[unmpo20]

plus C

[uccvertigo]

Why do you slip so many errors ? for (sin^2)(theta)=((x-3)/2)^2 you cannot simply take the square on both sides, you get something like sin(theta)=+OR-(x-3)/2;

The right way to do it was to substitute from the beggining sin(theta)=(x-3)/2, that way you would do it mathematically correct.

I know it doesn't influence the result, but you know error can cancel themself out and pass unnoticed, but if you were taking a test than it wouldn't be fully correct.

[Gsaurusrex]

AMAZING!

[KingHenryXI]

I've got a great Calc teacher, but you sir are convenient as fuck.

[mkcdcf]

OMG thank you SOooo much!!

[lilangel0072]

this got complicated when you tried to make it into a "nice clean answer">.< ...

[herojha]

sallu miya, great!

[herojha]

great!

[umeirdestination]

aw....man!!!

[MoGaDeX]

I wish my teacher explained things this well...

[godspeed28]

You're a wizard! <8-0

[hero3bash]

holy shit.. i feel so stupid

[andreiduffy]

11:50 there is sin θ * cos θ. not sinθ+cosθ..

ಠ_ಠ

[andreiduffy]

You have unbelievable concentration ability...

[Djole0]

dont understand why u first cancel those 2 from the ( x-3)/2 and then in antoehr line multiply with 2/2... and before u done that u could write sqrt( 1- (x-3)^2 /4) as sqrt(4 - ( x-3)^2)... ?

[05321]

@Djole0 He cancelled the 2 from under the (x-3)/2 because it was originally multiplied by 2. He then added the 2/2 to make get a root(4) in there. adding a 2/2 doesn't do anything because 2/2 = 1, so multiplied by 1 doesn't change anything.

[archieparchie1]

he forgot to add plus c at the end!

[hsktoadftw]

damnnnnn ur a beast

[rinwhr]

Damn...

[Wurstdesmonats]

nice one ^^

[bboydjoe]

instead of doin all that after 13:15, you could just draw a right triangle with angle theta and use SOHCAHTOA to get the required angles

[alexgordonepic]

your hand hurts and my brain hurts. i need a break before the next one.

I appreciate the help.

[abdy1345]

you forgot + C at the end

[taurus50592]

applause

[phillywilly666]

what if you chose to use sin^2x=1-cos^2x would the answer be different

[WildChildftw]

Maybe you will get a different indefinite integral, but it is just as correct as the other way around.

[mrperfectprashant]

zynot91210  sin2x is 2sinx cosx

[zynot91210]

Hey man, at 11:53 its

sin2x=(sinx cosx + cosx sinx) not (sinx cosx + cosx + sinx)

[thevidfather]

Ah well, it was an inadvertent error. He didn't carry it forward or anything.

[zynot91210]

Yea, just saying incase someone gets confused.