Z sin x. cos xdx (2) Z x 3 e x dx (3) Z x 3 ln |x|dx

Let's go through each of these integrals one by one.

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### Integral 1: \( \int z \sin x \cos x \, dx \)

To integrate \( z \sin x \cos x \) with respect to \(x\), we assume that \(z\) is a constant with respect to \(x\). This allows us to factor \(z\) out of the integral:

\[
\int z \sin x \cos x \, dx = z \int \sin x \cos x \, dx
\]

#### Step 1: Simplify \( \sin x \cos x \)
We can use the double-angle identity for sine:
\[
\sin x \cos x = \frac{1}{2} \sin(2x)
\]

Thus, the integral becomes:
\[
z \int \sin x \cos x \, dx = z \int \frac{1}{2} \sin(2x) \, dx = \frac{z}{2} \int \sin(2x) \, dx
\]

#### Step 2: Integrate \( \sin(2x) \)
The integral of \( \sin(2x) \) is:
\[
\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)
\]

Therefore,
\[
\frac{z}{2} \int \sin(2x) \, dx = \frac{z}{2} \left(-\frac{1}{2} \cos(2x)\right) = -\frac{z}{4} \cos(2x) + C
\]

### Final Answer:
\[
\int z \sin x \cos x \, dx = -\frac{z}{4} \cos(2x) + C
\]

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### Integral 2: \( \int x^3 e^x \, dx \)

To solve this integral, we will use **integration by parts** multiple times. For integration by parts, recall the formula:
\[
\int u \, dv = uv - \int v \, du
\]

#### Step 1: Set up Integration by Parts
Let:
- \( u = x^3 \) (so that \( du = 3x^2 \, dx \))
- \( dv = e^x \, dx \) (so that \( v = e^x \))

Using the integration by parts formula, we get:
\[
\int x^3 e^x \, dx = x^3 e^x - \int 3x^2 e^x \, dx
\]

#### Step 2: Apply Integration by Parts Again on \( \int 3x^2 e^x \, dx \)
We need to repeat the process for \( \int 3x^2 e^x \, dx \):
\[
\int 3x^2 e^x \, dx = 3 \left(x^2 e^x - \int 2x e^x \, dx \right)
\]

#### Step 3: Apply Integration by Parts Again on \( \int 2x e^x \, dx \)
We apply integration by parts one more time:
\[
\int 2x e^x \, dx = 2 \left(x e^x - \int e^x \, dx\right) = 2 \left(x e^x - e^x\right) = 2x e^x - 2e^x
\]

#### Putting It All Together
Combining all the parts, we have:
\[
\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C
\]

### Final Answer:
\[
\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C
\]

---

### Integral 3: \( \int x^3 \ln |x| \, dx \)

For this integral, we’ll also use **integration by parts**. 

#### Step 1: Set up Integration by Parts
Let:
- \( u = \ln |x| \) (so that \( du = \frac{1}{x} \, dx \))
- \( dv = x^3 \, dx \) (so that \( v = \frac{x^4}{4} \))

Using the integration by parts formula:
\[
\int x^3 \ln |x| \, dx = \frac{x^4}{4} \ln |x| - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx
\]

#### Step 2: Simplify the Remaining Integral
The remaining integral simplifies as follows:
\[
\int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \int \frac{x^3}{4} \, dx = \frac{1}{4} \int x^3 \, dx
\]

#### Step 3: Integrate \( x^3 \)
\[
\int x^3 \, dx = \frac{x^4}{4}
\]
So,
\[
\frac{1}{4} \int x^3 \, dx = \frac{1}{4} \cdot \frac{x^4}{4} = \frac{x^4}{16}
\]

### Final Answer:
\[
\int x^3 \ln |x| \, dx = \frac{x^4}{4} \ln |x| - \frac{x^4}{16} + C
\]

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Would you like further clarification on any part of these calculations?

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Here are some further questions to consider:
1. What if we used \(x^2\) instead of \(x^3\) in the second integral?
2. How would we integrate \( \int x e^x \, dx \) using integration by parts?
3. How do we handle \(\int \sin x \cos x \, dx\) without assuming a constant factor?
4. Can we solve \(\int e^x \ln |x| \, dx\) using a similar approach?
5. How does the constant \(z\) affect the first integral in applications?

**Tip:** For complex integrals, integration by parts can be used repeatedly to break down the problem into manageable steps.

Detailed solutions for integrals involving sin(x)cos(x), x^3e^x, and x^3ln|x|. Learn integration techniques including integration by parts and trigonometric identities. Suitable for students in Calculus I or II.

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