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Sagot :
Sure, let's solve the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex].
To solve this integral, we will use integration by parts. Integration by parts is given by the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we'll let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \cos 3x \, dx \][/tex]
Now, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \cos 3x \, dx = \frac{1}{3} \sin 3x \][/tex]
Now, we apply the integration by parts formula:
[tex]\[ \int e^{2x} \cos 3x \, dx = e^{2x} \cdot \frac{1}{3} \sin 3x - \int \left( \frac{1}{3} \sin 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \int e^{2x} \sin 3x \, dx \][/tex]
Now we need to solve the integral [tex]\(\int e^{2x} \sin 3x \, dx\)[/tex]. We can use integration by parts again for this integral. Let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \sin 3x \, dx \][/tex]
Again, find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \][/tex]
Applying integration by parts formula:
[tex]\[ \int e^{2x} \sin 3x \, dx = e^{2x} \cdot \left(-\frac{1}{3} \cos 3x\right) - \int \left(-\frac{1}{3} \cos 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos 3x \, dx \][/tex]
Now let's denote [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] as [tex]\(I\)[/tex]. Then from the above steps:
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \left( -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} I \right) \][/tex]
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x - \frac{4}{9} I \][/tex]
Combine the [tex]\(I\)[/tex] terms on one side:
[tex]\[ I + \frac{4}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \left( 1 + \frac{4}{9} \right) I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \frac{13}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
Solve for [tex]\(I\)[/tex]:
[tex]\[ I = \frac{9}{13} \left( \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \right) \][/tex]
[tex]\[ I = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x \][/tex]
Hence, the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] is:
[tex]\[ \int e^{2x} \cos 3x \, dx = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x + C \][/tex]
Where [tex]\(C\)[/tex] is the constant of integration.
To solve this integral, we will use integration by parts. Integration by parts is given by the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Here, we'll let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \cos 3x \, dx \][/tex]
Now, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \cos 3x \, dx = \frac{1}{3} \sin 3x \][/tex]
Now, we apply the integration by parts formula:
[tex]\[ \int e^{2x} \cos 3x \, dx = e^{2x} \cdot \frac{1}{3} \sin 3x - \int \left( \frac{1}{3} \sin 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \int e^{2x} \sin 3x \, dx \][/tex]
Now we need to solve the integral [tex]\(\int e^{2x} \sin 3x \, dx\)[/tex]. We can use integration by parts again for this integral. Let:
[tex]\[ u = e^{2x} \quad \text{and} \quad dv = \sin 3x \, dx \][/tex]
Again, find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ du = \frac{d}{dx}(e^{2x}) \, dx = 2e^{2x} \, dx \][/tex]
[tex]\[ v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \][/tex]
Applying integration by parts formula:
[tex]\[ \int e^{2x} \sin 3x \, dx = e^{2x} \cdot \left(-\frac{1}{3} \cos 3x\right) - \int \left(-\frac{1}{3} \cos 3x \right) \cdot 2e^{2x} \, dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos 3x \, dx \][/tex]
Now let's denote [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] as [tex]\(I\)[/tex]. Then from the above steps:
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \left( -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} I \right) \][/tex]
[tex]\[ I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x - \frac{4}{9} I \][/tex]
Combine the [tex]\(I\)[/tex] terms on one side:
[tex]\[ I + \frac{4}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \left( 1 + \frac{4}{9} \right) I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
[tex]\[ \frac{13}{9} I = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \][/tex]
Solve for [tex]\(I\)[/tex]:
[tex]\[ I = \frac{9}{13} \left( \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x \right) \][/tex]
[tex]\[ I = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x \][/tex]
Hence, the integral [tex]\(\int e^{2x} \cos 3x \, dx\)[/tex] is:
[tex]\[ \int e^{2x} \cos 3x \, dx = \frac{3}{13} e^{2x} \sin 3x + \frac{2}{13} e^{2x} \cos 3x + C \][/tex]
Where [tex]\(C\)[/tex] is the constant of integration.
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