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Sagot :
To find the antiderivative [tex]\( F(t) \)[/tex] that satisfies the given condition [tex]\( F\left(\frac{\pi}{4}\right) = 5 \)[/tex], we will follow these steps:
1. Find the general antiderivative of [tex]\( f(t) = \csc^2(t) \)[/tex]:
The function [tex]\( \csc^2(t) \)[/tex] is the derivative of [tex]\(-\cot(t)\)[/tex]. Hence, the indefinite integral (antiderivative) of [tex]\( \csc^2(t) \)[/tex] is:
[tex]\[ \int \csc^2(t) \, dt = -\cot(t) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
2. Define the general form of [tex]\( F(t) \)[/tex]:
Thus, the general antiderivative [tex]\( F(t) \)[/tex] can be expressed as:
[tex]\[ F(t) = -\cot(t) + C \][/tex]
3. Apply the initial condition to find [tex]\( C \)[/tex]:
We are given that [tex]\( F\left(\frac{\pi}{4}\right) = 5 \)[/tex]. Plugging [tex]\( t = \frac{\pi}{4} \)[/tex] into the antiderivative:
[tex]\[ F\left(\frac{\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right) + C \][/tex]
Recall that [tex]\( \cot\left(\frac{\pi}{4}\right) = 1 \)[/tex]:
[tex]\[ -1 + C = 5 \][/tex]
Solving for [tex]\( C \)[/tex]:
[tex]\[ C = 5 + 1 = 6 \][/tex]
4. Write the particular solution with the constant [tex]\( C \)[/tex]:
Substituting [tex]\( C = 6 \)[/tex] back into our expression for [tex]\( F(t) \)[/tex]:
[tex]\[ F(t) = -\cot(t) + 6 \][/tex]
Thus, the antiderivative [tex]\( F(t) \)[/tex] that satisfies the given condition is:
[tex]\[ F(t) = 6 - \frac{\cos(t)}{\sin(t)} \][/tex]
1. Find the general antiderivative of [tex]\( f(t) = \csc^2(t) \)[/tex]:
The function [tex]\( \csc^2(t) \)[/tex] is the derivative of [tex]\(-\cot(t)\)[/tex]. Hence, the indefinite integral (antiderivative) of [tex]\( \csc^2(t) \)[/tex] is:
[tex]\[ \int \csc^2(t) \, dt = -\cot(t) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
2. Define the general form of [tex]\( F(t) \)[/tex]:
Thus, the general antiderivative [tex]\( F(t) \)[/tex] can be expressed as:
[tex]\[ F(t) = -\cot(t) + C \][/tex]
3. Apply the initial condition to find [tex]\( C \)[/tex]:
We are given that [tex]\( F\left(\frac{\pi}{4}\right) = 5 \)[/tex]. Plugging [tex]\( t = \frac{\pi}{4} \)[/tex] into the antiderivative:
[tex]\[ F\left(\frac{\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right) + C \][/tex]
Recall that [tex]\( \cot\left(\frac{\pi}{4}\right) = 1 \)[/tex]:
[tex]\[ -1 + C = 5 \][/tex]
Solving for [tex]\( C \)[/tex]:
[tex]\[ C = 5 + 1 = 6 \][/tex]
4. Write the particular solution with the constant [tex]\( C \)[/tex]:
Substituting [tex]\( C = 6 \)[/tex] back into our expression for [tex]\( F(t) \)[/tex]:
[tex]\[ F(t) = -\cot(t) + 6 \][/tex]
Thus, the antiderivative [tex]\( F(t) \)[/tex] that satisfies the given condition is:
[tex]\[ F(t) = 6 - \frac{\cos(t)}{\sin(t)} \][/tex]
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