Answer:
[tex]f(theta)=sin(theta) - cos(theta)[/tex] + C
This is my first time doing a double integral, so im only 90% sure in my answer
Step-by-step explanation:
You pretty much want to take the double integral of sinx + cosx
The anti-derivative of sinx = -cosx
The anti-derivative of cosx = sinx
So f' = -cosx + sinx
Now lets take the integral of f':
The anti-derivative of -cosx = sinx
The anti-derivative of sinx = -cosx
So, f(x) = sinx - cosx
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Work Shown:
I'll use x in place of theta since its easier to type on a keyboard.
f '' (x) = sin(x) + cos(x)
f ' (x) = -cos(x) + sin(x) + C ..... integrate both sides; dont forget the plus C
f ' (0) = 1
f ' (0) = -cos(0) + sin(0) + C
-cos(0) + sin(0) + C = 1
-1 + 0 + C = 1
C = 1+1
C = 2
So,
f ' (x) = -cos(x) + sin(x) + C
turns into
f ' (x) = -cos(x) + sin(x) + 2
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Now integrate both sides of the first derivative to get the original f(x) function
f ' (x) = -cos(x) + sin(x) + 2
f(x) = -sin(x) - cos(x) + 2x + D .... apply integral; D is some constant
f(0) = -sin(0) - cos(0) + 2(0) + D
f(0) = 0 - 1 + 0 + D
f(0) = D - 1
f(0) = 2
D-1 = 2
D = 2+1
D = 3
We have f(x) = -sin(x) - cos(x) + 2x + D update to f(x) = -sin(x) - cos(x) + 2x + 3
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So f '' (x) = sin(x) + cos(x) becomes f(x) = -sin(x) - cos(x) + 2x + 3 when f(0) = 2 and f ' (0) = 1
The last step is to replace every x with theta so that we get back to the original variable.
f(x) = -sin(x) - cos(x) + 2x + 3 turns into f(θ) = -sin(θ) - cos(θ) + 2θ + 3
