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

Explanation

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

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]

In this case, we set \( a = \pi \). Thus, we can write:

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

Using the trigonometric identities, we know:

\[ \sin(\pi - x) = \sin x \quad \text{and} \quad \cos(\pi - x) = -\cos x \]

However, since we are squaring the cosine term, we have:

\[ \cos^2(\pi - x) = \cos^2 x \]

Thus, substituting these into the integral gives us:

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

Now, we can expand this:

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

This simplifies to:

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

Adding \( I \) to both sides results in:

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

Thus, we can express \( I \) as:

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

Next, we need to evaluate the integral \( \int_{0}^{\pi} \sin x \cos^2 x \, dx \). We can use the substitution \( t = \cos x \), which gives \( dt = -\sin x \, dx \). The limits change as follows:

- When \( x = 0 \), \( t = \cos(0) = 1 \)
- When \( x = \pi \), \( t = \cos(\pi) = -1 \)

Thus, the integral becomes:

\[ \int_{0}^{\pi} \sin x \cos^2 x \, dx = -\int_{1}^{-1} t^2 \, dt = \int_{-1}^{1} t^2 \, dt \]

Now, we compute:

\[ \int_{-1}^{1} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

Substituting this back into our expression for \( I \):

\[ I = \frac{\pi}{2} \cdot \frac{2}{3} = \frac{\pi}{3} \]

Thus, the final result is:

\[ \boxed{\frac{\pi}{3}} \]