Integrate : `int e^x sin x dx.`

Answer

To solve the integral \( \int e^x \sin x \, dx \), we will use the method of integration by parts. Let's go through the steps in detail.

### Step 1: Set up the integral
Let:
\[ I = \int e^x \sin x \, dx \]

### Step 2: Apply integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]

Here, we can choose:
- \( u = \sin x \) (thus \( du = \cos x \, dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

Now, applying the integration by parts:
\[ I = e^x \sin x - \int e^x \cos x \, dx \]

### Step 3: Solve the new integral
Let’s denote the new integral as:
\[ J = \int e^x \cos x \, dx \]

Now, we will apply integration by parts again on \( J \):
Choose:
- \( u = \cos x \) (thus \( du = -\sin x \, dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

Applying integration by parts again:
\[ J = e^x \cos x - \int e^x (-\sin x) \, dx \]
\[ J = e^x \cos x + \int e^x \sin x \, dx \]
\[ J = e^x \cos x + I \]

### Step 4: Substitute back
Now substitute \( J \) back into the equation for \( I \):
\[ I = e^x \sin x - J \]
Substituting for \( J \):
\[ I = e^x \sin x - (e^x \cos x + I) \]
\[ I = e^x \sin x - e^x \cos x - I \]

### Step 5: Solve for \( I \)
Now, we can add \( I \) to both sides:
\[ 2I = e^x \sin x - e^x \cos x \]
\[ I = \frac{1}{2} (e^x \sin x - e^x \cos x) \]

### Step 6: Final answer
Thus, the integral is:
\[ I = \frac{e^x}{2} (\sin x - \cos x) + C \]
where \( C \) is the constant of integration.

### Summary of the solution:
\[ \int e^x \sin x \, dx = \frac{e^x}{2} (\sin x - \cos x) + C \]