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

Options

A. 0
B. 2
C. `pi`
D. none of these

Answer

To solve the integral

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

we can break it down into separate integrals:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx. \]

### Step 1: Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \, dx\)

The function \(x^2\) is an even function, so we can use the property of even functions:

\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx. \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \, dx = 2 \int_{0}^{\frac{\pi}{2}} x^2 \, dx. \]

Calculating the integral:

\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{from } 0 \text{ to } \frac{\pi}{2} = \frac{(\frac{\pi}{2})^3}{3} - 0 = \frac{\pi^3}{24}. \]

So,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \, dx = 2 \cdot \frac{\pi^3}{24} = \frac{\pi^3}{12}. \]

### Step 2: Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x \, dx\)

The function \(x \cos x\) is an odd function because:

\[ f(-x) = -x \cos(-x) = -x \cos x = -f(x). \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x \, dx = 0. \]

### Step 3: Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x \, dx\)

The function \(\tan^5 x\) is also an odd function:

\[ f(-x) = \tan^5(-x) = -\tan^5(x) = -f(x). \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x \, dx = 0. \]

### Step 4: Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx\)

This integral is straightforward:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi. \]

### Final Calculation

Now, we can combine all the results:

\[ I = \frac{\pi^3}{12} + 0 + 0 + \pi = \frac{\pi^3}{12} + \pi. \]

Thus, the final value of the integral is:

\[ \boxed{\frac{\pi^3}{12} + \pi}. \]