The value of the integral `I=int_(0)^((pi)/(2))(cosx-sinx)/(10-x^(2)+(pix)/(2))dx` is equal to

Options

A. `(pi)/(2)`
B. `pi`
C. 0
D. `4pi`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{10 - x^2 + \frac{\pi x}{2}} \, dx, \]

we can use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \), thus \( a + b = \frac{\pi}{2} \).

### Step 1: Apply the property of definite integrals

We can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos\left(\frac{\pi}{2} - x\right) - \sin\left(\frac{\pi}{2} - x\right)}{10 - \left(\frac{\pi}{2} - x\right)^2 + \frac{\pi}{2} \left(\frac{\pi}{2} - x\right)} \, dx. \]

### Step 2: Simplify the trigonometric functions

Using the identities \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \) and \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), we can rewrite the numerator:

\[ \cos\left(\frac{\pi}{2} - x\right) - \sin\left(\frac{\pi}{2} - x\right) = \sin x - \cos x. \]

### Step 3: Substitute and simplify the denominator

Now, we need to simplify the denominator:

\[ 10 - \left(\frac{\pi}{2} - x\right)^2 + \frac{\pi}{2} \left(\frac{\pi}{2} - x\right). \]

Calculating \( \left(\frac{\pi}{2} - x\right)^2 \):

\[ \left(\frac{\pi}{2} - x\right)^2 = \frac{\pi^2}{4} - \pi x + x^2. \]

Now substituting this into the denominator:

\[ 10 - \left(\frac{\pi^2}{4} - \pi x + x^2\right) + \frac{\pi^2}{4} - \frac{\pi x}{2}. \]

This simplifies to:

\[ 10 - \frac{\pi^2}{4} + \pi x - x^2 - \frac{\pi^2}{4} + \frac{\pi x}{2} = 10 - \frac{\pi^2}{2} + \frac{3\pi x}{2} - x^2. \]

### Step 4: Combine the integrals

Now, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{10 - \frac{\pi^2}{2} + \frac{3\pi x}{2} - x^2} \, dx. \]

### Step 5: Combine both integrals

Now we can combine both expressions for \( I \):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{10 - x^2 + \frac{\pi x}{2}} \, dx = -I. \]

### Step 6: Solve for \( I \)

Adding both sides gives:

\[ 2I = 0 \implies I = 0. \]

Thus, the value of the integral is:

\[ \boxed{0}. \]