If In=∫0π4tannxdx , then limn→∞n(In+In−2)=

In=∫0π4tannxdx=∫0π4tann−2⋅tan2xdx                               =∫0π4tann−2x(sec2x −1)dx                               =∫0π4tann−2xd(tanx)−In−2⇒In+In−2=∫0π4tann−2xd(tanx)=tann−1xn−1|0π4=1n−1          ∴ltn→∞n(In+In−2)=ltn→∞nn−1=1