revising for test - need help on qns[URL=http://imageshack.us][IMG]http://img107.imag (1 Viewer)

[ronaldinho]

show working please

thanks!

 

[watatank]

Re: revising for test - need help on qnshttp://img107.[/b]

lim (x--> 0) { 2sin2x/x }

= lim (x--> 0) {2* 2 sin2x/2x }

= 4 * lim (x--> 0) {sin2x/2x }

=4 *1

=4

 

[ronaldinho]

Re: revising for test - need help on qnshttp://img107.[/b]

[quote=watatank]lim (x--> 0) { 2sin2x/x }

= lim (x--> 0) {2* 2 sin2x/2x }

= 4 * lim (x--> 0) {sin2x/2x }

=4 *1

=4[/quote]

how did you get sin2x/2x = 1?

 

[LoneShadow]

Re: revising for test - need help on qnshttp://img107.[/b]

Integral[2dx/Sqrt(1-9x^2)]; sub x = Sin[@]/3. Then dx = Cos[@]d@/3, Sqrt[1-9x^2] = Sqrt[Cos^2[@]] = Cos[@]. you can work out the rest.


Integral[Sin^2[x/2]dx]; use the property that Sin^2[nx] = 0.5(1-Cos[2nx]); so Sin^[x/2] = 0.5(1-Cos[x]). you can work out the rest.


To express Sqrt[3]*Cos[x] - Sin[x] in the form of RCos[x+a], there's a formula in your text book. here's how to do it without the formula:

Sqrt[3]*Cos[x] - Sin[x] = RCos[x+a] = R{Cos[x]Cos[a]-Sin[x]Sin[a]} = {RCos[a]}Cos[x] - {RSin[a]}Sin[x]

so equating like terms: we must have RCos[a] = Sqrt[3] and RSin[a] = 1; Now u can draw the triangle with this properties and find Tan[a] = 1/Sqrt[3], which means a = pi/6. Now sub this into either RCos[a] = Sqrt[3] or RSin[a] = 1 to get R = 2.
So Sqrt[3]*Cos[x] - Sin[x] = 2Cos[x+pi/6]

solving the next bit is not difficult now.


for last question do what they say and u get Integral = Integral[udu/Sqrt[2-u^2]]. Now sub m=2-u^2 and it should work.

 

[LoneShadow]

Re: revising for test - need help on qnshttp://img107.[/b]

[QUOTE=ronaldinho]how did you get sin2x/2x = 1?[/QUOTE]

lim (x-->0) Sin[nx]/nx = 1. It should've been proved to u in school.

 

[ronaldinho]

Re: revising for test - need help on qnshttp://img107.[/b]

[quote=watatank]integral of [ x/3 * rt(x+2) dx ]

1/3* integral of [ x * (x+2)^-1 dx ]
[/quote]

hwod u get the 2nd step?

 

[watatank]

Re: revising for test - need help on qnshttp://img107.[/b]

Hmmm sorry about that...not sure what I was doing there....thats just wrong...terrible working out/typos...hang on il correct it....

integral of [ x/3 * rt(x+2) dx ]

1/3* integral of [ x * (x+2)^1/2 dx ]

u = x + 2
x = u-2
du = dx

I = 1/3 * integral of [ (u-2) * u^1/2 du ]

= 1/3 * integral of [ (u^(3/2) - 2u^(1/2) du ]

= 1/3 * [ 2/5*(u^(5/2) - 2/3*2u^(3/2)]

= 1/3 * u^(3/2)* [ 2/5*u - 1/3]

Fingers crossed that's fixed once and for all...