`int_(0)^(pi//2) x sin x cos x dx`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x \sin x \cos x \, dx \), we can use the identity for sine:

\[ \sin 2x = 2 \sin x \cos x \implies \sin x \cos x = \frac{1}{2} \sin 2x \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} x \left(\frac{1}{2} \sin 2x\right) \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \sin 2x \, dx \]

Now, let’s denote \( J = \int_{0}^{\frac{\pi}{2}} x \sin 2x \, dx \). Therefore, we have:

\[ I = \frac{1}{2} J \]

Next, we will solve \( J \) using integration by parts. We choose:

- \( u = x \) (thus \( du = dx \))
- \( dv = \sin 2x \, dx \) (thus \( v = -\frac{1}{2} \cos 2x \))

Using integration by parts, we have:

\[ J = \left[ u v \right]_{0}^{\frac{\pi}{2}} - \int v \, du \]

Substituting the values:

\[ J = \left[ x \left(-\frac{1}{2} \cos 2x\right) \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \left(-\frac{1}{2} \cos 2x\right) \, dx \]

Calculating the boundary term:

\[ \left[ -\frac{1}{2} x \cos 2x \right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \left(\frac{\pi}{2} \cos(\pi)\right) - \left(-\frac{1}{2} \cdot 0 \cdot \cos(0)\right) = -\frac{1}{2} \left(\frac{\pi}{2} \cdot (-1)\right) = \frac{\pi}{4} \]

Now, we need to compute the integral:

\[ \int_{0}^{\frac{\pi}{2}} -\frac{1}{2} \cos 2x \, dx = -\frac{1}{2} \cdot \left[ \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \cdot \left(0 - 0\right) = 0 \]

So we have:

\[ J = \frac{\pi}{4} + 0 = \frac{\pi}{4} \]

Finally, substituting back into our expression for \( I \):

\[ I = \frac{1}{2} J = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8} \]

Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{8}} \]