f'(sinx)=cos^2x,f'(1)=1 then f(x)=
http://www.wolframalpha.com/input/?i=cos^2%28x%29
remember to add the constant of integration to that solution.
To find the function f(x), we need to integrate its derivative f'(x) with respect to x. Let's begin by integrating f'(sin(x)) = cos^2(x).
Integrating f'(sin(x)) with respect to x, we can use the substitution method. Let u = sin(x), then du = cos(x) dx. Rearranging, dx = du / cos(x).
Replacing dx and sin(x) in terms of u, we have:
∫f'(sin(x)) dx = ∫(cos^2(x)) dx
∫f'(u) du/cos(x) = ∫(cos^2(x)) dx
Since cos^2(x) = 1/2(1 + cos(2x)), we can rewrite the equation as:
∫f'(u) du/cos(x) = ∫(1/2)(1 + cos(2x)) dx
Expanding the integral on the right-hand side:
∫f'(u) du/cos(x) = ∫(1/2) dx + ∫(1/2)cos(2x) dx
The first integral on the right-hand side is ∫(1/2) dx = 1/2 x + C1, where C1 is the constant of integration.
For the second integral, we can use the substitution method again. Let w = 2x, then dw = 2 dx. Rearranging, dx = dw / 2.
Replacing dx and cos(2x) in terms of w, we have:
∫(1/2)cos(2x) dx = ∫(1/2)cos(w) (dw/2) = (1/4) ∫cos(w) dw
Integrating (1/4) ∫cos(w) dw = (1/4) sin(w) + C2, where C2 is the constant of integration.
Substituting the integral back into the equation, we get:
∫f'(u) du/cos(x) = 1/2 x + (1/4) sin(w) + C1
Since u = sin(x) and w = 2x, we can replace them back into the equation:
∫f'(sin(x)) du/cos(x) = 1/2 x + (1/4) sin(2x) + C1
Finally, to solve for f(x), we can integrate both sides with respect to x:
∫f'(sin(x)) du/cos(x) dx = ∫ (1/2 x + (1/4) sin(2x) + C1) dx
Integrating the left-hand side:
∫f'(sin(x)) du/cos(x) dx = ∫ du
Since f'(1) = 1, the left-hand side simplifies to:
f(u) + C3 = u + C3
Replacing u with sin(x) and simplifying further, we obtain:
f(x) = sin(x) + C3
Therefore, the function f(x) is f(x) = sin(x) + C3, where C3 is the constant of integration.