Question

f'(sinx)=cos^2x,f'(1)=1 then f(x)=

Answer 1

http://www.wolframalpha.com/input/?i=cos^2%28x%29

Answer 2

remember to add the constant of integration to that solution.

Answer 3

[AI-generated answer — Explain Bot]

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.