If `A=int_0^pi cosx/(x+2)^2 \ dx`, then `int_0^(pi//2) (sin 2x)/(x+1) \ dx` is equal to

Options

A. `1/2+1/(pi+2)-A`
B. `1/(pi+2)-A`
C. `1+1/(pi+2)-A`
D. `A-1/2-1/(pi+2)`

Explanation

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{x+1} \, dx \), we will use a substitution and integration by parts. Let's go through the steps:

### Step 1: Substitution
Let \( t = 2x \). Then, \( dt = 2dx \) or \( dx = \frac{dt}{2} \). The limits of integration change as follows:
- When \( x = 0 \), \( t = 0 \).
- When \( x = \frac{\pi}{2} \), \( t = \pi \).

Thus, we can rewrite the integral:
\[ I = \int_0^{\pi} \frac{\sin t}{\frac{t}{2} + 1} \cdot \frac{dt}{2} \]
This simplifies to:
\[ I = \frac{1}{2} \int_0^{\pi} \frac{\sin t}{\frac{t}{2} + 1} \, dt \]

### Step 2: Simplifying the Integral
Now, we can rewrite the integral:
\[ I = \int_0^{\pi} \frac{\sin t}{t + 2} \, dt \]

### Step 3: Integration by Parts
Let:
- \( u = \sin t \) ⇒ \( du = \cos t \, dt \)
- \( dv = \frac{1}{t + 2} \, dt \) ⇒ \( v = \ln(t + 2) \)

Using integration by parts:
\[ I = \left[ \sin t \ln(t + 2) \right]_0^{\pi} - \int_0^{\pi} \ln(t + 2) \cos t \, dt \]

### Step 4: Evaluating the Boundary Terms
Now, we evaluate the boundary terms:
\[ \left[ \sin t \ln(t + 2) \right]_0^{\pi} = \sin(\pi) \ln(\pi + 2) - \sin(0) \ln(2) = 0 - 0 = 0 \]
Thus, we have:
\[ I = - \int_0^{\pi} \ln(t + 2) \cos t \, dt \]

### Step 5: Relating to Given Integral \( A \)
We know from the problem statement that:
\[ A = \int_0^{\pi} \frac{\cos x}{(x + 2)^2} \, dx \]
By changing the variable in the integral \( A \) to \( t = x \), we can see that:
\[ \int_0^{\pi} \frac{\cos t}{(t + 2)^2} \, dt = A \]

### Final Step: Conclusion
Thus, we can conclude that:
\[ I = -A \]

### Summary
The value of the integral \( \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{x + 1} \, dx \) is equal to \( -A \).