IIT JEE , AIEEE Forum - Mathematics , Trignometry

(**343**)

ans this plz...

matrishvan

quiet simple.This is an example of integration over differentiatiom.As the two operations are inverse with the outside operation being of integration,the question reduces to tan inverse(1) - taninverse(-1) = pi/4-(-pi/4) = pi/2.

Same thing for other integral,with the order reversed.

rate me as i need that!!

(**343**)

but since tan^-1x=cot^-1(1/x), say both be f(x),then why does the value of the integral of their derivatives, over the same limits change ?

matrishvan

you got it wrong.

taninv(1/x) is not always equal to cotinv(x)

try to put -1.taninv-1 is (-pi/4) for cotinv(-1) is( 3pi/4)

are you clear with this?

(**343**)

Let tan^-1x =

......(1).

So,

x = tan

.

(1/x) = cot

cot^-1(1/x) =

......(2).

Hence from (1) and (2), tan^-1x=cot^-1(1/x).

Then what's wrong in the above steps???

feynmann

It has to do with their principal values .

krishna.gopal

Perfect feynmann. See 12121212 if cot(-pi/4)=cot(3pi/4) we can't say that -pi/4 = 3pi/4. We will have to take principal values of taninverse and cotinverse

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