Question #37ae3

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Answer 1 · 2017-10-24T03:42:02

# 0.#

Suppose that, #I=int_0^oolnx/(1+x^2)dx.#

We subst. #x=tany.:. dx=sec^2ydy.#

Further, #x=0 rArr y=0,"&, as "x to oo, y to pi/2.#

#:. I=int_0^(pi/2) (lntany/cancel(1+tan^2y))*cancel(sec^2y)dy, i.e., #

# I=int_0^(pi/2)lntanydy.........(star^1).#

Now, we know that, #int_0^af(y)dx=int_0^af(a-y)dx....(R).#

We apply the Result R to #(star^1)"with "a=pi/2, &, f(y)=lntany.#

#:. I=int_0^(pi/2)lntan(pi/2-y)dy, or, #

# I=int_0^(pi/2)lncotydy...................(star^2).#

#:. (star^1)+(star^2) rArr 2I=int_0^(pi/2)lntanydy+int_0^(pi/2)lncotydy,#

#:. 2I=int_0^(pi/2)[lntany+lncoty}dy,#

#=int_0^(pi/2)[ln(tany*coty)]dy,#

#=int_0^(pi/2)ln1dy,#

#:. 2I=int_0^(pi/2)0dy=0.#

# rArr I=0.#

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