Evaluate:`int_(-pi/2)^(pi/2)log((a-s intheta)/(a+sintheta))dtheta,a >0

1.	Evaluate:`int_(-pi/2)^(pi/2)log((a-s intheta)/(a+sintheta))dtheta,a >0`
Answer» `f(theta)=log((a-sintheta)/(a+sin theta))` `:.f(-theta)=log((a+sin theta)/(a-sin theta))` `=-log((a-sin theta)/(a+sin theta))` `=-f(theta)` Hence the integrand is anodd function. So, the given integral is zero.

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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