How to calculate this? #lim_(n->oo)int_(1/n)^n(arctanx^2)/(1+x^2)dx#.

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Answer 1 · 2017-05-22T02:05:51

#I=int(arctanx)^2/(1+x^2)dx#

Letting #u=arctanx# so #du=1/(1+x^2)dx#:

#I=intu^2du=1/3u^3=1/3(arctanx)^3+C#

So:

#int_(1/n)^n(arctanx)^2/(1+x^2)dx=1/3(arctan(n))^3-1/3(arctan(1/n))^3#

Note that #lim_(nrarroo)arctan(n)=pi/2# and #lim_(nrarr0)1/n=0#:

#lim_(nrarroo)int_(1/n)^n(arctanx)^2/(1+x^2)dx=1/3(pi/2)^3-1/3(arctan(0))^3#

#=1/3(pi^3/8)=pi^3/24#