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Mathematics: Post your doubts here!

haha101

haha101

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haha . Look if two vectors are collinear of parallel then they have a common ratio between them . suppose a=(2.4.6) and b=(1.2.3)

a/b =2 and so they are both collinear
 
PANDA-

PANDA-

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haha101 said:
haha . Look if two vectors are collinear of parallel then they have a common ratio between them . suppose a=(2.4.6) and b=(1.2.3)

a/b =2 and so they are both collinear
Click to expand...

OR... Their unit vectors are equal.
 
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!!
 
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