Trivial Q 2 (1 Viewer)

[OLDMAN]

Give Spice Girl's integral notation a run, btw where's Spice?

I{0--->pi/2}f(x)dx integral upper limit pi/2, lower limit 0 wrt x, and f(x)=1/(1+tan^.25 (x))

 

[Affinity]

hmm Pi/4 ,

wasn't that <u>trivial</u>

very nice question~

 

[underthesun]

how do you figure this out?

seems like im missing out on the "trick"

 

[ezzy85]

is this similar to the sinx/sinx + cosx question? where you make 2 identities and add them, then divide by 2.

 

[freaking_out]

Originally posted by Affinity
hmm Pi/4 ,

wasn't that <u>trivial</u>

very nice question~

what trivial?????? ....don't even have a clue as to where to start with this question.

Give Spice Girl's integral notation a run, btw where's Spice?

yeah, spice girl's hanging around more in the chemistry forum these days.

 

[underthesun]

I would have done I{a-->b}f(x)dx = I{a-->b}f(a-x)dx

or something like that? I quite forgot, have been slacking off integration..

 

[ezzy85]

Originally posted by underthesun
I would have done I{a-->b}f(x)dx = I{a-->b}f(a-x)dx

or something like that? I quite forgot, have been slacking off integration..

ye, thats what i was thinking. using cotx for f(a-x)

 

[Affinity]

well.. alternatively, you could do a x=arctan(y^4) substitution.

 

[OLDMAN]

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Affinity quote: well.. alternatively, you could do a x=arctan(y^4) substitution.
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Did you get the right answer pi/4 using that substitution ? Didn't think of that. Tell us, Affinity, where did you come from? Mars?!

 

[freaking_out]

well what substitution did u have in mind oldman?

 

[OLDMAN]

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Originally posted by underthesun
I would have done I{a-->b}f(x)dx = I{a-->b}f(a-x)dx

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ezzy85 quote:
ye, thats what i was thinking. using cotx for f(a-x)
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didn't use substitution, but used above special integration rule.

 

[freaking_out]

Originally posted by OLDMAN
...Tell us, Affinity, where did you come from? Mars?!

yep, your prediction fits in well with the fact that mars is closest to earth in 60 thousand years...

 

[OLDMAN]

Just a note : .25 is actually irrelevant, it could be any number eg.

I{0--->pi/2}(1/(1+tan^.1(x))dx =I{0--->pi/2}(1/(1+cot^.1(x))dx

thus 2I=I{0--->pi/2}(1)dx giving pi/4.

 

[freaking_out]

Originally posted by OLDMAN

I{0--->pi/2}(1/(1+tan^.1(x))dx =I{0--->pi/2}(1/(1+cot^.1(x))dx

thus 2I=I{0--->pi/2}(1)dx giving pi/4.

hey, can u tell me how u got from the first step 2 the second??

 

[OLDMAN]

freaking_out:

1/(1+tan^.1(x)) + 1/(1+cot^.1(x)) =1

 

[freaking_out]

Originally posted by OLDMAN
freaking_out:

1/(1+tan^.1(x)) + 1/(1+cot^.1(x)) =1

i'm sorry, but i still don't get it...(i'm braindead from the trials)...

anyway, did u use any trig identities???? how did u get them to equal 2 one??

 

[ND]

Just write it down and do it. It would come out to (2+tan^.1(x)+cot^.1(x))/(2+tan^.1(x)+cot^.1(x))=1

 

[freaking_out]

Originally posted by ND
Just write it down and do it. It would come out to (2+tan^.1(x)
+cot^.1(x))/(2+tan^.1(x)+cot^.1(x))=1

thanks. .....faar out.... i soo lazy!!!!

 

[OLDMAN]

or
1/(1+tan^.1(x)) + 1/(1+cot^.1(x)) =1/(1+tan^.1(x)) + tan^.1(x)/(1+tan^.1(x)) =1 since cot=1/tan.

Quite useful to remember 1/(1+tan)+1/(1+cot)=1