To solve the integral of $e^{-x} \cos(7x) \, dx$ using integration by parts and a reduction formula, we proceed step by step.

Step 1: Set up the integral and use integration by parts

We need to evaluate the integral:

$I = \int e^{-x} \cos(7x) \, dx$

Using the integration by parts formula:

$\int u \, dv = u v - \int v \, du$

We can choose:

• $u = \cos(7x)$, so $du = -7 \sin(7x) \, dx$
• $dv = e^{-x} \, dx$, so $v = -e^{-x}$

Step 2: Apply integration by parts

Using the formula:

$I = -e^{-x} \cos(7x) + 7 \int e^{-x} \sin(7x) \, dx$

Let the new integral be:

$I_1 = \int e^{-x} \sin(7x) \, dx$

Step 3: Use integration by parts again on $I_1$

We apply integration by parts again to $I_1$.

This time, choose:

• $u = \sin(7x)$, so $du = 7 \cos(7x) \, dx$
• $dv = e^{-x} \, dx$, so $v = -e^{-x}$

Applying integration by parts:

$I_1 = -e^{-x} \sin(7x) - 7 \int e^{-x} \cos(7x) \, dx$

Step 4: Solve the system of equations

We now have two expressions for our integrals:

1. $I = -e^{-x} \cos(7x) + 7 I_1$
2. $I_1 = -e^{-x} \sin(7x) - 7 I$

Substitute the expression for $I_1$ from equation 2 into equation 1:

$I = -e^{-x} \cos(7x) + 7 \left( -e^{-x} \sin(7x) - 7 I \right)$

Simplifying:

$I = -e^{-x} \cos(7x) - 7 e^{-x} \sin(7x) - 49 I$

Now, collect all terms involving $I$ on one side:

$I + 49 I = -e^{-x} (\cos(7x) + 7 \sin(7x))$

$50 I = -e^{-x} (\cos(7x) + 7 \sin(7x))$

Step 5: Solve for $I$

Finally, divide both sides by 50:

$I = -\frac{1}{50} e^{-x} (\cos(7x) + 7 \sin(7x)) + C$

where $C$ is the constant of integration.

Final Answer:

$\int e^{-x} \cos(7x) \, dx = -\frac{1}{50} e^{-x} (\cos(7x) + 7 \sin(7x)) + C$

Would you like more details, or do you have any questions?

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Related Questions:

1. How do we use integration by parts in other trigonometric-exponential integrals?
2. What is the general reduction formula for integrals involving $e^{ax} \cos(bx)$?
3. How can the technique be extended to $\int e^{ax} \sin(bx) \, dx$?
4. Can we apply a similar approach to integrals with higher powers of $x$ in the exponent?
5. How do we solve integrals involving products of two trigonometric functions?

Tip: In complex integrals, always look for patterns that allow you to simplify the system or use symmetry.