Answer
a)$\frac{1}{2}$
b)$\frac{-1}{2}$
Work Step by Step
a)Let u=$\tan x$=> du=$sec^2xdx$
x=0=>u=0,x=$\frac{\pi}{4}=>u=1$
=$\int ^{\pi/4}_0tan xsec^2xdx$
=$\int^1_0udu$
=$[\frac{u^2}{2}]^1_0$
=$\frac{1^2}{2}-0$
=$\frac{1}{2}$
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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
=$\frac{-1}{2}$
