1∫cotx dx=∫cosxsinxdx
let u=sinx
therefore du=cosx dx
→∫cotx dx=∫duu
→ln|u|+c
→ln|sinx|+c from 0-pi
→ans=∞
this sum came in AIEEE 2009..
∫[cotx]dx from 0 to pi where [.]denotes greatest integer fnctn
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13 Answers
1
11@debosmit-
A hint..
[x]+[-x]=-1 (x is not a whole number)
where [.] greatest integer function..;)
1but when x equal to pi/4,cotx is a whole number.so can u apply this and integrate?i`ve tried this method but i hv this doubt....
11I=∫(0 to pi) [Cot x]
Now cot(pi-x)=-cot x
Apprlying property of definite integral: ∫(0 to pi) cot x=∫(0 to pi) cot(pi-x)
Again, [-x]=-1-[x]
Thus,
I=-I-∫(0 to pi)dx
Therefore, 2 I =- pi
or, I= -pi/2
1yes it is.. but my qstn is can u apply this property at all points?
11Yeah why not!
The property is actually back integration.Instead of adding from 0 to pi we add from pi to 0!
11Okay at pi/4 cot x =1(integer).[At 3 pi/4 also integer]
[x]=-1-[x] only when x is not an integer. But I don't think that the answer of the integration differs due to that.Check from graph..
62Area of the graph will not chaneg if u remove finite nnumber of points
It will not change if u remove all rational points as well.. (infact as long as u remove countable or finite number of points.. There is slightly more but lets not get into that)
So The area will not change
Remember the rationals are far far far fewer than the irrationals...
1so still v`l nt get the exact integral value....actually it is almost negligible