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Expression on the Image:

$\int \left( x \sin(x^2) + 2x^2 \cos(x^2) - 3x \right) dx$

Solution:

This integral can be broken down into three parts:

1. $\int x \sin(x^2) \, dx$
2. $\int 2x^2 \cos(x^2) \, dx$
3. $\int -3x \, dx$

1. Solve $\int x \sin(x^2) \, dx$

Let $u = x^2$, then $du = 2x \, dx$.

This means: $\frac{1}{2} \int \sin(u) \, du = -\frac{1}{2} \cos(u) + C_1 = -\frac{1}{2} \cos(x^2) + C_1$

2. Solve $\int 2x^2 \cos(x^2) \, dx$

We perform integration by parts, using the formula: $\int u \, dv = uv - \int v \, du$ Let:

• $u = x^2$, then $du = 2x \, dx$
• $dv = \cos(x^2) dx$

To integrate $dv = \cos(x^2) dx$ directly, is complicated, so this part will be approximated by referring to advanced calculus or numerical methods.

3. Solve $\int -3x \, dx$

This is straightforward: $\int -3x \, dx = -\frac{3}{2}x^2 + C_2$

Combining Results:

Putting everything together:

$-\frac{1}{2} \cos(x^2) + \text{(integral of the second part)} - \frac{3}{2}x^2 + C$

For a complete answer, an approximation or special techniques are needed for the second integral.

Would you like details on any specific part of the solution? Here are some further questions that can explore this topic:

1. How can integration by parts be applied to the second integral?
2. What techniques can be used to approximate the integral of $\cos(x^2)$?
3. Why is substitution useful in integrals like the first one?
4. How do different integration techniques apply to complex integrals?
5. Can numerical methods provide a more precise result for the second integral?

Tip: When working with integrals involving trigonometric functions and polynomials, substitution and integration by parts are key techniques to master.