If `A= int _(0)^(pi) (sin x)/(x ^(2))dx,` then `int _(0)^(pi//2) (cos 2 x )/(x) dx` is equal to:

Options

A. `1-A`
B. `3/2-A`
C. `A-1`
D. `1+A`

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos(2x)}{x} \, dx \]

given that

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

### Step 1: Substitution
We start by making the substitution \( t = 2x \). Then, differentiating both sides gives us:

\[ dt = 2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{2}. \]

### Step 2: Change of Limits
Next, we change the limits of integration. When \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{2} \), \( t = \pi \). Thus, the integral becomes:

\[ I = \int_{0}^{\pi} \frac{\cos(t)}{\frac{t}{2}} \cdot \frac{dt}{2} = \int_{0}^{\pi} \frac{\cos(t)}{t} \, dt. \]

### Step 3: Integration by Parts
Now we will use integration by parts on the integral \( \int_{0}^{\pi} \frac{\cos(t)}{t} \, dt \). We set:

- \( u = \frac{1}{t} \) so that \( du = -\frac{1}{t^2} \, dt \)
- \( dv = \cos(t) \, dt \) so that \( v = \sin(t) \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:

\[ I = \left[ \frac{\sin(t)}{t} \right]_{0}^{\pi} - \int_{0}^{\pi} \sin(t) \left(-\frac{1}{t^2}\right) dt. \]

### Step 4: Evaluate the Boundary Terms
Evaluating the boundary terms:

\[ \left[ \frac{\sin(t)}{t} \right]_{0}^{\pi} = \frac{\sin(\pi)}{\pi} - \lim_{t \to 0} \frac{\sin(t)}{t} = 0 - 1 = -1. \]

### Step 5: Substitute Back
Now substituting back into the equation for \( I \):

\[ I = -1 + \int_{0}^{\pi} \frac{\sin(t)}{t^2} \, dt. \]

### Step 6: Recognize the Integral
Notice that the integral \( \int_{0}^{\pi} \frac{\sin(t)}{t^2} \, dt \) is exactly \( A \). Thus, we can write:

\[ I = -1 + A. \]

### Final Answer
Therefore, we conclude that:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\cos(2x)}{x} \, dx = A - 1. \]