The integral displayed is:

$\int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x - \frac{1}{4} \int e^{2x} \cos x \, dx$

This equation is derived using integration by parts. Let me provide a step-by-step explanation of how this equation is constructed:

Step 1: Integration by Parts Formula

The formula for integration by parts is:

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

Step 2: Assign $u$ and $dv$

Here, the given integral is $\int e^{2x} \cos x \, dx$.

• Let $u = \cos x$, so $du = -\sin x \, dx$.
• Let $dv = e^{2x} \, dx$, so $v = \frac{1}{2} e^{2x}$ (since $\int e^{2x} \, dx = \frac{1}{2} e^{2x}$).

Step 3: Apply Integration by Parts

Using the formula:

$\int e^{2x} \cos x \, dx = \left( \frac{1}{2} e^{2x} \cos x \right) - \int \left( \frac{1}{2} e^{2x} (-\sin x) \, dx \right)$

$= \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x \, dx$

Step 4: Repeat for $\int e^{2x} \sin x \, dx$

For the new integral, $\int e^{2x} \sin x \, dx$, apply integration by parts again:

• Let $u = \sin x$, so $du = \cos x \, dx$.
• Let $dv = e^{2x} \, dx$, so $v = \frac{1}{2} e^{2x}$.

Applying the formula again:

$\int e^{2x} \sin x \, dx = \left( \frac{1}{2} e^{2x} \sin x \right) - \int \left( \frac{1}{2} e^{2x} \cos x \, dx \right)$

$= \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \int e^{2x} \cos x \, dx$

Step 5: Substitution and Solving

Substitute this result into the original equation:

$\int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x - \frac{1}{4} \int e^{2x} \cos x \, dx$

Combine terms to isolate $\int e^{2x} \cos x \, dx$:

$\int e^{2x} \cos x \, dx + \frac{1}{4} \int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x$

$\frac{5}{4} \int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x$

$\int e^{2x} \cos x \, dx = \frac{1}{5} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x \right)$

$\int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x - \frac{1}{4} \int e^{2x} \cos x \, dx$

This matches the result shown in your image.

Would you like me to simplify further or explain any specific part in detail?

5 Questions You Could Explore:

1. How is integration by parts useful for solving these types of integrals?
2. Can this method be applied to other combinations of exponentials and trigonometric functions?
3. What are the properties of integrals involving $e^{ax}$ multiplied by trigonometric functions?
4. How would the solution change if the exponential were $e^{3x}$ instead of $e^{2x}$?
5. How can we use this solution to evaluate definite integrals?

Tip:

Integration by parts often involves recursive patterns; keep track of repeated integrals to simplify your calculations.