`int_(-2)^(2) X sin^(10) x dx =?`

Answer

To solve the integral \( \int_{-2}^{2} x \sin^{10}(x) \, dx \), we can follow these steps:

### Step 1: Identify the Function
Let \( f(x) = x \sin^{10}(x) \).

### Step 2: Check if the Function is Odd or Even
To determine if \( f(x) \) is odd or even, we need to evaluate \( f(-x) \):
\[ f(-x) = -x \sin^{10}(-x) \]
Using the property of sine, \( \sin(-x) = -\sin(x) \), we can rewrite this as:
\[ f(-x) = -x (-\sin(x))^{10} = -x \sin^{10}(x) \]
Thus, we have:
\[ f(-x) = -f(x) \]
This shows that \( f(x) \) is an odd function.

### Step 3: Use the Property of Integrals of Odd Functions
The property of integrals states that if \( f(x) \) is an odd function, then:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
for any \( a \).

### Step 4: Apply the Property
Since our function \( f(x) = x \sin^{10}(x) \) is odd, we can apply this property:
\[ \int_{-2}^{2} x \sin^{10}(x) \, dx = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{-2}^{2} x \sin^{10}(x) \, dx = 0 \]

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