Math, asked by anwesh2000, 1 year ago

integrate limits 1/2 to 1 (1/x*cosec^101 ( x - 1/x)) dx

Answers

Answered by shubhi9643
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..
shubhi9643: that 1/x= t and this will be cancelled with 1/t from -1/t^2
anwesh2000: ohh thankyou so much for ur kind help..
Answered by Sidyandex
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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