The following integral `int_(pi/4)^(pi/2)(2cosecx)^(17)` dx is equal toA. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du`B. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`C. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`D. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du`

The following integral `int_(pi/4)^(pi/2)(2cosecx)^(17)` dx is equal t

1.	The following integral `int_(pi/4)^(pi/2)(2cosecx)^(17)` dx is equal toA. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du`B. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`C. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`D. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du`
Answer» Correct Answer - A

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)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`