S8.7.2.1 AP calculus Exam (typo problem)

[karush]

ok I thot this was just observation to get b.

but maybe not

I saw some rather hefty substations to get different answers

 

[skeeter]

if it were $\displaystyle \int \dfrac{\cos{t}}{1+\sin{t}} \, dt$, then choice (b) would fit

$\dfrac{1}{1+\sin{t}} \cdot \dfrac{1-\sin{t}}{1-\sin{t}} = \dfrac{1-\sin{t}}{1-\sin^2{t}} = \dfrac{1-\sin{t}}{\cos^2{t}} = \dfrac{1}{\cos^2{t}} - \dfrac{1}{\cos{t}} \cdot \dfrac{\sin{t}}{\cos{t}} = \sec^2{t} - \sec{t}\tan{t}$

$\displaystyle \int \sec^2{t} - \sec{t}\tan{t} \, dt = \tan{t} - \sec{t} + C$

methinks choice (a) has a typo

 

[karush]

$\dfrac{d}{dx}\left(\tan \theta \:+\sec \theta \right)
=-\dfrac{1}{-1+\sin(x)}=\dfrac{1}{1-\sin(x)}$

ok I did this to one of the answers but when I integrated it it did not return the

so something is missing :(