∫
−1
1
​
sin(x)⋅
2
1
​
(3x
2
−1)dx

The given integral is:

\[
\int_{-1}^{1} \sin(x) \cdot \frac{2}{1} (3x^2 - 1) \, dx
\]

We can simplify the expression first:

\[
\int_{-1}^{1} \sin(x) \cdot 2 (3x^2 - 1) \, dx
\]

### Key observation: 
This is an integral over a symmetric interval \([-1, 1]\), and it's important to determine if the integrand is odd or even. 

- The function \(\sin(x)\) is **odd** (i.e., \(\sin(-x) = -\sin(x)\)).
- The polynomial \(3x^2 - 1\) is **even** (i.e., \(3(-x)^2 - 1 = 3x^2 - 1\)).

Multiplying an odd function by an even function results in an **odd function**. The integral of any odd function over a symmetric interval like \([-1, 1]\) is **zero**.

Thus, we conclude:

\[
\int_{-1}^{1} \sin(x) \cdot 2 (3x^2 - 1) \, dx = 0
\]

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How do you identify if a function is odd or even?
2. What happens if the interval is not symmetric for odd functions?
3. How can you compute the integral of an even function over a symmetric interval?
4. How do you handle integrals involving trigonometric functions?
5. What are the properties of definite integrals over symmetric intervals?

**Tip:** Always check for symmetry (odd/even nature) before solving definite integrals over symmetric intervals—it can often simplify your work!

Learn how to solve the definite integral ∫−1 to 1 sin(x)⋅2(3x²−1)dx using properties of odd and even functions. This example demonstrates how recognizing the symmetry of a function can simplify the solution. Suitable for university-level calculus students.

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