`int_(0)^(pi//2) x sinx cos x dx=?`

Options

A. `(pi)/(4)`
B. `(pi)/(8)`
C. `(pi)/(12)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x \sin x \cos x \, dx \), we will use the property of integration that states:

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

### Step 1: Apply the property of integration

Let \( a = 0 \) and \( b = \frac{\pi}{2} \). Therefore, we can write:

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

### Step 2: Simplify the integrand

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

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

### Step 3: Expand the integral

Now, we can expand this integral:

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

Notice that the second integral on the right side is again \( I \):

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

### Step 4: Rearranging the equation

Now, we can rearrange the equation to isolate \( I \):

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

### Step 5: Evaluate the integral \( \int_{0}^{\frac{\pi}{2}} \cos x \sin x \, dx \)

To evaluate \( \int_{0}^{\frac{\pi}{2}} \cos x \sin x \, dx \), we can use the substitution \( u = \sin x \), then \( du = \cos x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{2} \), which corresponds to \( u = 0 \) to \( u = 1 \):

\[ \int_{0}^{\frac{\pi}{2}} \cos x \sin x \, dx = \int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2} \]

### Step 6: Substitute back into the equation for \( I \)

Now substituting back into our equation for \( I \):

\[ 2I = \frac{\pi}{2} \cdot \frac{1}{2} \]

\[ 2I = \frac{\pi}{4} \]

### Step 7: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = \frac{\pi}{8} \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} x \sin x \cos x \, dx = \frac{\pi}{8} \]

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