Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 68

$4-\pi$

See figure. The integration will be over $dy$. We have $A=\int_{-\pi/4}^{\pi/4}(tan^2y-(-tan^2y)dy =\int_{-\pi/4}^{\pi/4}(2tan^2y)dy=2\int_{-\pi/4}^{\pi/4}(sec^2y-1)dy =2\ tan(y)|_{-\pi/4}^{\pi/4}-2y|_{-\pi/4}^{\pi/4} =2\ (tan(\pi/4)-tan(-\pi/4))-2(\pi/4-(-\pi/4))=4-\pi$