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3 ) f(x)=sin(2x+3) F ( x ) = − c o s ( a x + b ) ...

Creato il: 28 novembre 2025

Risposto usando GPT-5.1 Thinking da Chat01

Domanda

3
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f(x)=sin(2x+3)

F
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−
c
o
s
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a
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b
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k
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F(x)=−
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cos(ax+b)
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Pensiero

Clarifying integration problem

It looks like the user shared a math problem related to finding an antiderivative. They likely need to determine the constants a, b, and k such that the derivative of F(x) matches f(x) = sin(2x+3). The original antiderivative form is F(x) = -cos(ax+b)/k + C, though formatting confusion occurred while pasting the problem.

Solving the integral and finding constants

It seems the user wants the antiderivative of sin(2x+3) expressed as F(x) = -k cos(ax+b) + C, and to find the values for a, b, and k. For ∫sin(2x+3) dx, using u = 2x+3, the result is -1/2 cos(2x+3) + C, where a=2, b=3, and k = 1/2. If their form involves division by k, k=2 instead.

Analyzing two possible forms of the solution

The user might be dealing with two potential forms of the antiderivative for f(x) = sin(2x+3). For one form, F(x) = -cos(ax+b)/k + C, we get a

Thought for 16s

Risposta

For

f(x)=sin⁡(2x+3)f(x)=\sin(2x+3)

an antiderivative is

F(x)=∫sin⁡(2x+3) dx=−12cos⁡(2x+3)+C.F(x)=\int \sin(2x+3)\,dx = -\frac12\cos(2x+3)+C.

Matching this to your parameter form:

  • If

    F(x)=−cos⁡(ax+b)k+C,F(x) = -\frac{\cos(ax+b)}{k} + C,

    then

    a=2,b=3,k=2.a = 2,\quad b = 3,\quad k = 2.

  • If

    F(x)=−kcos⁡(ax+b)+C,F(x) = -k\cos(ax+b) + C,

    then

    a=2,b=3,k=12.a = 2,\quad b = 3,\quad k = \tfrac12.

Either way, the actual antiderivative is F(x)=−12cos⁡(2x+3)+CF(x) = -\frac12\cos(2x+3)+C.

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