Evaluate `int_(1)^(5)sqrt(x-2)sqrt(x-1)dx`.

1.	Evaluate `int_(1)^(5)sqrt(x-2)sqrt(x-1)dx`.
Answer» Correct Answer - `2` `int_(1)^(5)sqrt(x-2sqrt(x-1))dx=int_(1)^(5)sqrt((sqrt(x-1)-1)^(2))dx` `=int_(1)^(5)\|sqrt(x-1)-1\|dx` `=int_(1)^(2)(1-sqrt(x-1))dx+int_(2)^(5)(sqrt(x-1)-1)dx` `=\|x-(2x(x-1)^(3//2))/3\|_(1)^(2)+\|(2(x-1)^(3//2))/3-x\|_(2)^(5)` `=(2-2/3-1+0)+(16/3-5-2/3+2)=2`

Discussion

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
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Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
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