Question

Find f. (antiderivative)
f '(t) = 4 cos (t) + sec^2(t), −π/2 < t < π/2, f(π/3) = 3

Answer 1

I keep getting 4sint+tan t -3.4641

Answer 2

\[4 \int\limits \cos(t) dt + \int\limits \sec^2(t) dt = 4 \sin(t) + \tan(t) + C\]
Then:
\[4 \sin(\frac{\pi}{3}) + \tan( \frac{\pi}{3}) + C = 3 \implies C= 3 - 2 \sqrt3 - \sqrt 3 \implies C = 3 - \sqrt3 \]
\[\therefore f(t)= 4\sin(t) + \tan(t) + 3 - \sqrt3\]

Answer 3

I got sec^2 +1

Answer 4

Don't use calculators, exact numbers when given nice angles like:
\[\frac{\pi}{3}, \frac{\pi}{6}, \frac{\pi}{4}\]

Answer 5

It didnt accept 4sin(t)+tan(t)+3−sqrt 3

Answer 6

I'm sorry, it should be:
\[4 \sin(t) + \tan(t) + 3 - 3\sqrt3\]

Answer 7

I ran out of guesses but thanks anyway