$Q6(paper1)$

$(a)$ Given that $>1,$ use the substitution $u=\frac{1}{x}to$ show that

${\displaystyle \underset{1}{\overset{}{}}}\frac{1}{1+x^2}dx={\displaystyle \underset{\frac{1}{}}{\overset{1}{}}}\frac{1}{1+u^2}du$

$(b)$ Hence show that $arctan+arctan(\frac{1}{})=\frac{}{2}.$