Find f(x)= on ((-pi/2), (pi/2)) when f'(x)= 4+ tan^2x and f(0)= 4?

Answer 1

f'(x) = 4 + tan^2 x = 3 + sec^2 x
The trig identity is actually
1 + tan^2 x = sec^2 x

f(x) = 3x + tanx + 4
where 4 = constant of i ntegration that satisfies f(0) = 4

Answer 2

df/dx = 4 + tan^2x

Use the trig identity: tan^2x = 1 - sec^2x

So df = (4 + 1 - sec^2x)dx = (5 - sec2x)dx

Integrate to get f = 5x - tan(x) + C since Sec2x integrated is just tan(x)

Now set x=0 and get f(0) = 4 = 0 - 0 + C or C = 4

So the function is: f(x) = 5x - tan(x) + 4

The limitation of restricting the function to -pi/2 to pi/2 is because the tan function goes to +/- infinity at these values

Answer 3

easy..the answer is 2.543688879..i think
wait maybe im wrong idk!