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Sagot :
Certainly! Let's solve the integral step by step:
Given:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx \][/tex]
First, expand the integrand:
[tex]\[ \left( e^{a x} - e^{-a x} \right)^2 = (e^{a x})^2 - 2 e^{a x} e^{-a x} + (e^{-a x})^2 \][/tex]
Simplify the terms:
[tex]\[ (e^{a x})^2 = e^{2 a x} \quad \text{and} \quad (e^{-a x})^2 = e^{-2 a x} \quad \text{and} \quad e^{a x} e^{-a x} = e^{0} = 1 \][/tex]
Therefore:
[tex]\[ \left( e^{a x} - e^{-a x} \right)^2 = e^{2 a x} - 2 \cdot 1 + e^{-2 a x} = e^{2 a x} + e^{-2 a x} - 2 \][/tex]
Now our integral is:
[tex]\[ \int \left( e^{2 a x} + e^{-2 a x} - 2 \right) \, dx \][/tex]
We can split this integral into three separate integrals:
[tex]\[ \int e^{2 a x} \, dx + \int e^{-2 a x} \, dx - \int 2 \, dx \][/tex]
Compute each of these integrals separately:
1. For the first integral:
[tex]\[ \int e^{2 a x} \, dx \][/tex]
Let [tex]\( u = 2 a x \)[/tex]. Then [tex]\( du = 2 a \, dx \)[/tex], and hence:
[tex]\[ dx = \frac{du}{2a} \][/tex]
Thus,
[tex]\[ \int e^{2 a x} \, dx = \int e^u \cdot \frac{du}{2a} = \frac{1}{2a} \int e^u \, du = \frac{1}{2a} e^u + C = \frac{1}{2a} e^{2 a x} \][/tex]
2. For the second integral:
[tex]\[ \int e^{-2 a x} \, dx \][/tex]
Let [tex]\( v = -2 a x \)[/tex]. Then [tex]\( dv = -2 a \, dx \)[/tex], and hence:
[tex]\[ dx = \frac{dv}{-2a} \][/tex]
Thus,
[tex]\[ \int e^{-2 a x} \, dx = \int e^v \cdot \frac{dv}{-2a} = \frac{1}{-2a} \int e^v \, dv = \frac{1}{-2a} e^v + C = \frac{1}{-2a} e^{-2 a x} \][/tex]
3. For the third integral:
[tex]\[ \int 2 \, dx = 2x + C \][/tex]
Combining these results, we get:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} e^{2 a x} + \frac{1}{-2a} e^{-2 a x} - 2x + C \][/tex]
Simplify the coefficients:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} e^{2 a x} - \frac{1}{2a} e^{-2 a x} - 2x + C \][/tex]
Factor out [tex]\(\frac{1}{2a}\)[/tex]:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} \left( e^{2 a x} - e^{-2 a x} \right) - 2x + C \][/tex]
Finally, we combine this result into the final answer:
[tex]\[ -2x + \text{Piecewise}\left( \frac{2a e^{2 a x} - 2a e^{-2 a x}}{4 a^2}, a \neq 0 \right) + \text{Piecewise}\left( 2x, \text{True} \right) + C \][/tex]
The simplified result:
[tex]\[ -2x + \text{Piecewise}\left( \left( \frac{e^{2 a x} - e^{-2 a x}}{2a}, a \neq 0 \right), (2x, \text{True})\right) + C \][/tex]
This matches the final result given.
Given:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx \][/tex]
First, expand the integrand:
[tex]\[ \left( e^{a x} - e^{-a x} \right)^2 = (e^{a x})^2 - 2 e^{a x} e^{-a x} + (e^{-a x})^2 \][/tex]
Simplify the terms:
[tex]\[ (e^{a x})^2 = e^{2 a x} \quad \text{and} \quad (e^{-a x})^2 = e^{-2 a x} \quad \text{and} \quad e^{a x} e^{-a x} = e^{0} = 1 \][/tex]
Therefore:
[tex]\[ \left( e^{a x} - e^{-a x} \right)^2 = e^{2 a x} - 2 \cdot 1 + e^{-2 a x} = e^{2 a x} + e^{-2 a x} - 2 \][/tex]
Now our integral is:
[tex]\[ \int \left( e^{2 a x} + e^{-2 a x} - 2 \right) \, dx \][/tex]
We can split this integral into three separate integrals:
[tex]\[ \int e^{2 a x} \, dx + \int e^{-2 a x} \, dx - \int 2 \, dx \][/tex]
Compute each of these integrals separately:
1. For the first integral:
[tex]\[ \int e^{2 a x} \, dx \][/tex]
Let [tex]\( u = 2 a x \)[/tex]. Then [tex]\( du = 2 a \, dx \)[/tex], and hence:
[tex]\[ dx = \frac{du}{2a} \][/tex]
Thus,
[tex]\[ \int e^{2 a x} \, dx = \int e^u \cdot \frac{du}{2a} = \frac{1}{2a} \int e^u \, du = \frac{1}{2a} e^u + C = \frac{1}{2a} e^{2 a x} \][/tex]
2. For the second integral:
[tex]\[ \int e^{-2 a x} \, dx \][/tex]
Let [tex]\( v = -2 a x \)[/tex]. Then [tex]\( dv = -2 a \, dx \)[/tex], and hence:
[tex]\[ dx = \frac{dv}{-2a} \][/tex]
Thus,
[tex]\[ \int e^{-2 a x} \, dx = \int e^v \cdot \frac{dv}{-2a} = \frac{1}{-2a} \int e^v \, dv = \frac{1}{-2a} e^v + C = \frac{1}{-2a} e^{-2 a x} \][/tex]
3. For the third integral:
[tex]\[ \int 2 \, dx = 2x + C \][/tex]
Combining these results, we get:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} e^{2 a x} + \frac{1}{-2a} e^{-2 a x} - 2x + C \][/tex]
Simplify the coefficients:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} e^{2 a x} - \frac{1}{2a} e^{-2 a x} - 2x + C \][/tex]
Factor out [tex]\(\frac{1}{2a}\)[/tex]:
[tex]\[ \int \left( e^{a x} - e^{-a x} \right)^2 \, dx = \frac{1}{2a} \left( e^{2 a x} - e^{-2 a x} \right) - 2x + C \][/tex]
Finally, we combine this result into the final answer:
[tex]\[ -2x + \text{Piecewise}\left( \frac{2a e^{2 a x} - 2a e^{-2 a x}}{4 a^2}, a \neq 0 \right) + \text{Piecewise}\left( 2x, \text{True} \right) + C \][/tex]
The simplified result:
[tex]\[ -2x + \text{Piecewise}\left( \left( \frac{e^{2 a x} - e^{-2 a x}}{2a}, a \neq 0 \right), (2x, \text{True})\right) + C \][/tex]
This matches the final result given.
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