The value of `int_(-pi//2)^(pi//2) ( x^(5)+ x sin^(2)x +2 tan^(-1)x ) dx` is

Options

A. `pi`
B. 2
C. 1
D. 0

Answer

To solve the integral

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^5 + x \sin^2 x + 2 \tan^{-1} x \right) dx, \]

we will check if the integrand is an odd function.

### Step 1: Define the function
Let

\[ f(x) = x^5 + x \sin^2 x + 2 \tan^{-1} x. \]

### Step 2: Check if \(f(x)\) is an odd function
To check if \(f(x)\) is an odd function, we need to evaluate \(f(-x)\):

\[ f(-x) = (-x)^5 + (-x) \sin^2(-x) + 2 \tan^{-1}(-x). \]

### Step 3: Simplify \(f(-x)\)
Now, simplify each term:

1. \((-x)^5 = -x^5\) (odd function)
2. \(\sin^2(-x) = \sin^2 x\) (even function), thus \((-x) \sin^2(-x) = -x \sin^2 x\) (odd function)
3. \(\tan^{-1}(-x) = -\tan^{-1}(x)\) (odd function), thus \(2 \tan^{-1}(-x) = -2 \tan^{-1}(x)\) (odd function)

Putting it all together, we have:

\[ f(-x) = -x^5 - x \sin^2 x - 2 \tan^{-1} x = -\left( x^5 + x \sin^2 x + 2 \tan^{-1} x \right) = -f(x). \]

### Step 4: Conclude that \(f(x)\) is odd
Since \(f(-x) = -f(x)\), we conclude that \(f(x)\) is an odd function.

### Step 5: Evaluate the integral
The integral of an odd function over a symmetric interval \([-a, a]\) is zero:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^5 + x \sin^2 x + 2 \tan^{-1} x \right) dx = 0. \]

### Final Answer
The value of the integral is

\[ \boxed{0}. \]