integrate limits 1/2 to 1 (1/x*cosec^101 ( x - 1/x)) dx
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Answered by
4
Assume the limit to be equal to I
Take x=1/t
then dx= -1/t^2 dt
and cosec^101(1/t -t)= -cosec^101(t- 1/t)
so finally you'll get
I= -I
therefor I=0
anwesh2000:
sorry..but there is 1/x also outside..and after taking t you have not replaced..-1/t^2..you may be right but actually i didn't understand..so pls can u make me undestand..
that 1/x= t and this will be cancelled with 1/t from -1/t^2
ohh thankyou so much for ur kind help..
Answered by
1
We need to integrate the limits 1/2 to 1 (1/x*cosec^101 ( x - 1/x)) dx.
We can assume the limit to be equal to I. After that, we can take x=1/t.
It is essential to know that dx= -1/t^2 dt. The value of cosec^101(1/t -t) is -cosec^101(t- 1/t).
And hence we will get the value of I is –I.
Therefore the value of I is 0.
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