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integration Q (1 Viewer)

guesswho

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hey, can someone please help me with this Q;)
note: can't find the integration symbol, so this { symbol will have to do.

Question: using the result that the {f(x) dx = { f(a-x)dx,
show that { [ (1-sin2x) over (1+ sin2x)] dx = { (tan square x)dx.

note: from cambridge ex 5.6 Q2a.
 

OLDMAN

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rule is only valid for definite integral over a and 0. As I do not have the Cambridge book, I will work it backwards and get the limits.

Using (1-cos2x)/(1+cos2x) = tan^2(x) and cos2x=sin(pi/2-2x)

{(1-sin2x)/(1+sin2x)dx over unknown limit a and 0 =
{(1-sin2(a-x))/(1+sin2(a-x))dx={(1-sin(2a-2x))/(1+sin(2a-2x))dx

Therefore 2a=pi/2 and a=pi/4.
 

Mathematician

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Ur not really showing it, are u?

I see what u have done but the problem is:
show that { [ (1-sin2x) over (1+ sin2x)] dx = { (tan square x)dx.


what u did is u found a=pi/4 , now how does this help to show
[ (1-sin2x) over (1+ sin2x)]dx =tan^2xdx ?


guesswho ,note:(1-cos2x /(1+ cos2x) = tan^2(x)
since 2sin^2(x)=1-cos2x and 2cos^2(x)=1+cos2x
.
(u probably know this , lol).
 

spice girl

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dodgy question to start off with,

Mathematician, is this a definite or an indefinite integral?

If indefinite, I dun think the result is true at all...
 

OLDMAN

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guesswho must have forgotten to put the limits pi/4 and 0 in, so had to work them out. Anyway Mathematician's made the valid point that I skipped an important step. Question: any suggestions for typing definite integrals?
 

OLDMAN

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students might want to try this one out.
/pi/2
Evaluate l (sinx+.5)/(cosx+1+sinx)dx
/0
/a /a
Using l f(x)dx = l f(a-x)dx
/0 /0
 

OLDMAN

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That didn't seem to work. Try again. Evaluate

/pi/2
l (sinx+.5)/(sinx+1+cosx)dx
/0

using,
/a
l f(x)dx=
/0

/a
l f(a-x)dx
/0
 

guesswho

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Originally posted by OLDMAN
guesswho must have forgotten to put the limits pi/4 and 0 in, so had to work them out. Anyway Mathematician's made the valid point that I skipped an important step. Question: any suggestions for typing definite integrals?
sorry:eek: I did forget to put limits :p

Anyway, the limits from both the integrals are pi/4 and 0.


(hey, you were right!:eek: You're pretty smart :) )
 

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