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Mathematics: Post your doubts here!

littlecloud11

littlecloud11

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PhyZac said:
You are welcome, I am glad I did something to help you!!

Btw, ._. can you explain? like some steps are so weird to me.
Click to expand...

I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
Break tanⁿ⁺² x into tanⁿx* tan²x
I = ⌡( tanⁿx* tan²x + tanⁿ x) dx
I = ⌡( tanⁿx ( tan²x +1) sec²x
tan²x +1 = sec²x
I = ⌡( tanⁿxsec²x) dx

u = tan x
du = sec²x dx
dx = 1/sec²x du

Substitute tan x = u

I = ⌡ uⁿ xsec²x) *1/sec²x du
sec²x gets canceled.
I =⌡ uⁿ du
I = uⁿ⁺¹/(n + 1)

Change the limit of x for u and solve. Hope this helps.
 
daredevil

daredevil

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PANDA- said:
OR... Their unit vectors are equal.
Click to expand...
yeah that's correct too but that is kind of a solving the whole thing type method..... what he said is much simplar....
in other words what he is trying to say is that 2 vectors are parellel if they are like: - a=kb where a and b are vectors and k is a constant. =)
 
P

PhyZac

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littlecloud11 said:
I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
Break tanⁿ⁺² x into tanⁿx* tan²x
I = ⌡( tanⁿx* tan²x + tanⁿ x) dx
I = ⌡( tanⁿx ( tan²x +1) sec²x
tan²x +1 = sec²x
I = ⌡( tanⁿxsec²x) dx

u = tan x
du = sec²x dx
dx = 1/sec²x du

Substitute tan x = u

I = ⌡ uⁿ xsec²x) *1/sec²x du
sec²x gets canceled.
I =⌡ uⁿ du
I = uⁿ⁺¹/(n + 1)

Change the limit of x for u and solve. Hope this helps.
Click to expand...
OOOOhhh, Jazaki Allah Soooooooooo muuuchhhh!! Thank youuuuuuuuuuuu!!!!!!!!!!!!
May Alah S.W.T reward you, and give you the highest results, and have mercy on you and your family,Aameeeennn...
Thanks ALOT!!
 
daredevil

daredevil

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A star i have to go right now but i'll do it and post later today :)
i have done a bit of it actually :)
just the equations are complicated otherwise its just like any other such question :)
just do it with a clear head and fresh start and u'll be able to do it :)
 
MustafaMotani

MustafaMotani

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beeloooo said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf here you go anyone who can explain the last question as a whole please ?
Click to expand...
i) y=0
0=x(x-2)
x=0 and x=2
a=2 (since its visible from the graph that a cant be zero)

ii) b is a max pt.
dy/dx = 3x^2 -8x +4 =0
x= 2 and x= 2/3

b=2/3 (since both turning pts are visible from the graph and b is the first turning pt so i ll take the smaller value of x) (beside a was 2) :D
s
iii) integrate the function with limits 0 and 2

iv) It is asking for the minimum value of dy/dx... dy/dx= 3x^2 - 8x +4
thus yu just have to find minum value of that quadratic function
Two ways to do it .. Do completing squares or just derivate it again and equate it to zero .. u shall get x substitute that x to the dy/dx equation and that will be yur anser
 
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