Answer
$$\int_{-\pi/2}^0\Big(2+\tan\frac{t}{2}\Big)\sec^2\frac{t}{2}dt=3$$ $$\int_{-\pi/2}^{\pi/2}\Big(2+\tan\frac{t}{2}\Big)\sec^2\frac{t}{2}dt=8$$
Work Step by Step
Part a) $$A=\int_{-\pi/2}^0\Big(2+\tan\frac{t}{2}\Big)\sec^2\frac{t}{2}dt$$
We set $u=2+\tan\frac{t}{2}$, which means $$du=\frac{1}{2}\sec^2\frac{t}{2}dt$$ $$\sec^2\frac{t}{2}dt=2du$$
For $t=0$, we have $u=2+\tan0=2$
For $t=-\pi/2$, we have $u=2+\tan(-\pi/4)=2-1=1$
Therefore, $$A=2\int^2_1udu=2\times\frac{u^2}{2}\Big]^2_1=u^2\Big]^2_1$$ $$A=2^2-1=3$$
Part b) $$A=\int_{-\pi/2}^{\pi/2}\Big(2+\tan\frac{t}{2}\Big)\sec^2\frac{t}{2}dt$$
We set up the same substitution as in part a).
For $t=\pi/2$, we have $u=2+\tan(\pi/4)=2+1=3$
For $t=-\pi/2$, we have $u=2+\tan(-\pi/4)=2-1=1$
Therefore, $$A=2\int^3_1udu=u^2\Big]^3_1$$ $$A=3^2-1=8$$
