Let's solve each of the three integrals step by step.

Problem 1: $\int_0^\pi x \sin x \cos x \, dx$

This can be simplified using the trigonometric identity: $\sin x \cos x = \frac{1}{2} \sin(2x)$ Thus, the integral becomes: $\int_0^\pi x \sin x \cos x \, dx = \frac{1}{2} \int_0^\pi x \sin(2x) \, dx$

We can use integration by parts, where:

• Let $u = x$ and $dv = \sin(2x) dx$,
• So, $du = dx$ and $v = -\frac{1}{2} \cos(2x)$.

Now applying the integration by parts formula: $\int u \, dv = uv - \int v \, du$ $\frac{1}{2} \left[ -\frac{x}{2} \cos(2x) \right]_0^\pi + \frac{1}{2} \int_0^\pi \frac{1}{2} \cos(2x) \, dx$ Evaluating this integral gives: $I_1 = \boxed{0}$

Problem 2: $\int_{-1}^{1} |x| e^x \, dx$

The absolute value function splits the integral into two parts: $\int_{-1}^{1} |x| e^x \, dx = \int_{-1}^{0} (-x) e^x \, dx + \int_0^1 x e^x \, dx$

Now, compute each part:

1. For $\int_{-1}^{0} (-x) e^x \, dx$, we use integration by parts:

   • Let $u = -x$ and $dv = e^x dx$,
   • So, $du = -dx$ and $v = e^x$, $\int_{-1}^{0} (-x) e^x \, dx = -[(-x) e^x]_ {-1}^{0} + \int_{-1}^{0} e^x \, dx$ Evaluating gives: $\int_{-1}^{0} (-x) e^x \, dx = e^{-1}$

2. For $\int_0^1 x e^x \, dx$, we use the same process: $\int_0^1 x e^x \, dx = [xe^x]_0^1 - \int_0^1 e^x \, dx = e - 1$

Thus, the final result is: $I_2 = e - \frac{1}{e} - 1 = \boxed{e - 1 - \frac{1}{e}}$

Problem 3: $\int_{\frac{1}{e}}^e |\ln x| \, dx$

We need to split the integral based on the behavior of $|\ln x|$:

• For $x \in \left[\frac{1}{e}, 1\right]$, $\ln x \leq 0$, so $|\ln x| = -\ln x$,
• For $x \in [1, e]$, $\ln x \geq 0$, so $|\ln x| = \ln x$.

Thus, the integral becomes: $\int_{\frac{1}{e}}^e |\ln x| \, dx = \int_{\frac{1}{e}}^1 (-\ln x) \, dx + \int_1^e \ln x \, dx$

Now compute each part:

1. For $\int_{\frac{1}{e}}^1 (-\ln x) \, dx$, we use the formula $\int \ln x \, dx = x \ln x - x$: $\int_{\frac{1}{e}}^1 (-\ln x) \, dx = \left[ -x \ln x + x \right]_{\frac{1}{e}}^1 = 1 - \frac{1}{e}$

2. For $\int_1^e \ln x \, dx$, the same formula gives: $\int_1^e \ln x \, dx = \left[ x \ln x - x \right]_1^e = e - 1$

Thus, the final result is: $I_3 = 1 - \frac{1}{e} + e - 1 = \boxed{e - \frac{1}{e}}$

Would you like further details or clarifications? Here are some related questions to extend your understanding:

1. How does the trigonometric identity $\sin x \cos x = \frac{1}{2} \sin(2x)$ help simplify integrals?
2. What is the general approach to solving integrals that involve absolute value functions?
3. Why is integration by parts a useful technique for products of functions?
4. How does the integral $\int \ln x \, dx$ derive from the integration by parts method?
5. What are some applications of definite integrals in physics and engineering?

Tip: When faced with a product of two functions in an integral, consider using integration by parts!