Evaluate `int_(0)^(pi/4) sec^2x` `dx`

Explanation

`S=sum_(n=0)^(oo)(2^(n))/(a^(2n)+1)(agt1)`
`S_(n)=sum_(n=0)^(n)(2^(n))/(a^(2n)+1)`
`=(1)/(a+1)+(2)/(a^(2)+1)+(4)/(a^(4)+1)+(8)/(a^(8)+1)+"........"+(2^(n))/(a^(2^n)+1)`
`=(1)/(1+a)+(2)/(1+a^(2))+(4)/(1+a^(4))+(8)/(1+a^(8))+"........"+(2^(n))/(1+a^(2^n))`
`=(-(1)/(1-a)+(1)/(1-a))(1)/(1+a)+(2)/(1+a^(2))+(4)/(1+a^(4))+(8)/(1+a^(8))+"........"+(2^(n))/(1+a^(2^n))`
`=(1)/(a-1)+((1)/(1-a)+(1)/(1+a))+(2)/(1+a^(2))+(4)/(1+a^(4))+"........"+(2^(n))/(1+a^(2^n))`
`=(1)/(a-1)+((2)/(1-a^(2))+(2)/(1+a^(2)))+(4)/(1+a^(4))+"........"+(2^(n))/(1+a^(2^n))`
`" " vdots " "vdots " " vdots " "`
`S_(n)=(1)/(a-1)+(2^(n+1))/(1-a^(2^n+1))`
`S=lim_(n to oo)S_(n)=lim_(n to oo)((1)/(a-1)+(2^(n+1))/(1-a^(2^n+1)))`
`=lim_(n to oo)((1)/(a-1)+((2^(n+1))/(a^(2^n+1)))/((1)/(a^(2^n+1))-1))=(1)/(a-1)+(0)/(0-1)=(1)/(a-1)`.