April 1, 2015

Homework Help: calculus-integration!

Posted by Maria on Sunday, May 6, 2007 at 4:25pm.

should i use substitution?? if yes how should should i use it? plz i need some directions?

k plz someone? far i used trig. substitution. i got a=8, so i used x=asin()so according to this substitution i got x=8sin() and dx=8cos() d...then i used used the plug them in the original problem ...
so when i did that and simplfying it...i got 128sin^2()*8cos()/squarroot of 64-256sin^2...then how do i intergrate that? i am doing my best... am i doing it correct?

Are you certain the x in the numerator is squared? If it is not, then this helps...

divide numerator and denominator by 1/2

then you have INT x^2/sqrt(16-x^2)

Now, integrate it by parts, u= x^2, v= 1/sqrt(16-x^2) You will get power function, and an arcsin function.

integrate (2x^2)/squarroot of (64-4x^2)

no IT IS squared...shouldnt i be using trignometric substitution?

Yes, use substitution.
Let u = x^2
du = 2 x dx
x = sqrt u
dx = (1/2)/u^(1/2) du
and you get
(Integral of)[ 2u/sqrt(64 - 2u)](1/2)/u^(1/2) du
= (Integral of) sqrt(u/(64 - 2u) du

Letting v = 64 - 2u will simplify that further.

If it is not squared, yes, use trig substitution.

so integration by parts would help? or can i still use trignometric substittion?

i did use it...i got to the point of 128sin^2*8cos() d()/6-16sin() do i integrate this?

why would v=64-2u? becasue it is square root of 64-4x^ wouldnt v=8-2x? i am confused on that part...but plz so far i used trig. substitution...i got to the point 128sin^2*8cos d /8-16sin ? but i am not sure where to go after that in order to integrate it?...


i got the answer to be 2sin^2() that correct? i used trig. substitution

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