Question

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 1

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.