Find the mistake of the following evaluation of the integral`I=int_0^pi(dx)/(1+2sin^2x)``I=int_0^pi(dx)/(cos^2x+3sin^2x)``=int_0^pi(sec^2x dx)/(1+3tan^2x)=1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]pi0=0`

Find the mistake of the following evaluation of the integral`I=int_0^p

1.	Find the mistake of the following evaluation of the integral`I=int_0^pi(dx)/(1+2sin^2x)``I=int_0^pi(dx)/(cos^2x+3sin^2x)``=int_0^pi(sec^2x dx)/(1+3tan^2x)=1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]pi0=0`
Answer» Here the anti derivative `1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]=F(x)` is discontinuous at `x=pi//2` in the interval `[0,pi]`. Since `F((pi^(+))/2)=lim_(hto0)F((pi)/2+h)` `=lim_(hto0)(1/(sqrt(3)))tan^(-1){sqrt(3)"tan"(1/2pi+h)}` `=lim_(hto0)(1//sqrt(3))"tan"^(-1){-sqrt(3)coth}` `=(1/(sqrt(3))) tan^(-1)(-oo)=-pi//(2sqrt(3))` and `F(1/2pi-0)=pi//(2sqrt(3))!=F(1/2pi+0)` the second fundamental theorem of integral calculus is not applicable.

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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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4
`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
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))`