## Thomas' Calculus 13th Edition

a)$\frac{1}{2}$ b)$\frac{-1}{2}$
a)Let u=$\tan x$=> du=$sec^2xdx$ x=0=>u=0,x=$\frac{\pi}{4}=>u=1$ =$\int ^{\pi/4}_0tan xsec^2xdx$ =$\int^1_0udu$ =$[\frac{u^2}{2}]^1_0$ =$\frac{1^2}{2}-0$ =$\frac{1}{2}$ --- b)Use the same substitution as in part (a) ;x=$\frac{\pi}{4}=>u=-1, x=0=>u=0$ =$\int^0_{-\pi/4}\tan xsec^2xdx$ =$\int^0_{-1}udu$ =$[\frac{u^2}{2}]^0_{-1}$ =0-1/2 =$\frac{-1}{2}$