The integral shown in the image is:

$\int e^x \sin(x) \, dx$

This integral can be solved using the method of integration by parts. Let’s solve it step by step:

Step 1: Recall Integration by Parts Formula

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

Step 2: Select $u$ and $dv$

Let:

• $u = \sin(x)$ (so that $du = \cos(x) \, dx$)
• $dv = e^x \, dx$ (so that $v = e^x$, since $\int e^x \, dx = e^x$)

Step 3: Apply Integration by Parts

$\int e^x \sin(x) \, dx = e^x \sin(x) - \int e^x \cos(x) \, dx$

Step 4: Solve the Remaining Integral

Now, we need to compute $\int e^x \cos(x) \, dx$. Use integration by parts again:

• Let $u = \cos(x)$ (so that $du = -\sin(x) \, dx$)
• Let $dv = e^x \, dx$ (so that $v = e^x$)

$\int e^x \cos(x) \, dx = e^x \cos(x) - \int e^x (-\sin(x)) \, dx$ $\int e^x \cos(x) \, dx = e^x \cos(x) + \int e^x \sin(x) \, dx$

Step 5: Solve for the Original Integral

Let $I = \int e^x \sin(x) \, dx$. Substituting the result above, we have: $I = e^x \sin(x) - \left( e^x \cos(x) + I \right)$ $I = e^x \sin(x) - e^x \cos(x) - I$ $2I = e^x \sin(x) - e^x \cos(x)$ $I = \frac{1}{2} \left( e^x \sin(x) - e^x \cos(x) \right)$

Final Answer:

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

Where $C$ is the constant of integration.

Would you like further details or explanations on any of the steps?

Related Questions:

1. How does the choice of $u$ and $dv$ affect integration by parts?
2. Can this integral be solved using another method, such as substitution?
3. How does this solution extend to integrals of $e^x \cos(x)$?
4. What is the intuition behind using integration by parts for this problem?
5. How would the solution change if the integral were $\int e^{-x} \sin(x) \, dx$?

Tip:

When solving oscillatory integrals involving $e^x$ with trigonometric functions, expect to use integration by parts multiple times and solve for the original integral algebraically.