Find f.
f '(t) = 2 cos t + sec2t, −π/2 < t < π/2, f(π/3) = 3
im getting a wrong answer ...this is what i did...could someone help ?
f(t) = 2 sint + (1/2) ln(sec(2t) + tan(2t)) + C
given f(pi/3) = 3
so,
3 = 2 sin(pi/3) + (1/2) ln(sec(2pi/3) + tan(2pi/3)) + C
or
C = 2.945
so,
f(t) = 2 sint + (1/2) ln(sec(2t) + tan(2t)) + 2.945
f '(t) = 2 cos t + sec2t, −π/2 < t < π/2, f(π/3) = 3
im getting a wrong answer ...this is what i did...could someone help ?
f(t) = 2 sint + (1/2) ln(sec(2t) + tan(2t)) + C
given f(pi/3) = 3
so,
3 = 2 sin(pi/3) + (1/2) ln(sec(2pi/3) + tan(2pi/3)) + C
or
C = 2.945
so,
f(t) = 2 sint + (1/2) ln(sec(2t) + tan(2t)) + 2.945

You made some error in calculating C, but I can't tell why. Your f(t) is correct. Then
3 = 2sin(π/3) + ½lnlsec(2π/3) + tan(2π/3) + C ==>
C = 3  √(3)  ½ ln2  √(3) ≈ 0.6095.
Perhaps you punched something into a calculator incorrectly.
3 = 2sin(π/3) + ½lnlsec(2π/3) + tan(2π/3) + C ==>
C = 3  √(3)  ½ ln2  √(3) ≈ 0.6095.
Perhaps you punched something into a calculator incorrectly.