Help! calculus pls...?

Question

find f(x) if f"(x)=2 tanx sec^2(x) [secant squared of x] + 2 cosx and the rate of change of f with respect to x at the point P(π,0) is equal to 1

Answer 1

f''(x) = 2 tanx sec^2 x + 2 cosx

you must integrate on both sides

to integrate 2 tanx sec^2 x

put t= tan x

then dt = sec^2 x dx

thus

integral 2 tanx sec^2 x = integral 2t dt

= t^2

= tan^2 x

integral cos x is sinx

thus

f'(x) = tan^2 x + 2 sin x + C

C is the integration constant

at x=pi , f'(x) = 1 (given )

substituting

1 = 0 + 0 + C

thus C =1

f'(x) = tan^2 x + 2sin x + 1


f'(x) = sec^2 x - 1 + 2 sin x + 1 (since sec^2 x - tan^2 x =1)


f'(x) = sec^2 x + 2 sin x

integrating again

f(x) = tan x - 2cos x + C'

Answer 2

let u = tan(x) in the first integral, you'll find

du/dx = sec^2(x)