`int_(0)^(pi) xsin x cos^(4)x dx=`

Options

A. `(pi)/(10)`
B. `(pi)/(5)`
C. `-(pi)/(5)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{\pi} x \sin x \cos^4 x \, dx \), we can use the property of definite integrals that states:

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

### Step 1: Apply the property of definite integrals
We set \( a = \pi \) and rewrite the integral:

\[ I = \int_{0}^{\pi} x \sin x \cos^4 x \, dx = \int_{0}^{\pi} (\pi - x) \sin(\pi - x) \cos^4(\pi - x) \, dx \]

### Step 2: Simplify the expression
Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:

\[ I = \int_{0}^{\pi} (\pi - x) \sin x \cos^4 x \, dx \]

### Step 3: Expand the integral
Now, we can expand this integral:

\[ I = \int_{0}^{\pi} \pi \sin x \cos^4 x \, dx - \int_{0}^{\pi} x \sin x \cos^4 x \, dx \]

### Step 4: Combine the integrals
Let’s denote the second integral as \( I \):

\[ I = \pi \int_{0}^{\pi} \sin x \cos^4 x \, dx - I \]

### Step 5: Solve for \( I \)
Now, we can combine the terms:

\[ 2I = \pi \int_{0}^{\pi} \sin x \cos^4 x \, dx \]

Thus,

\[ I = \frac{\pi}{2} \int_{0}^{\pi} \sin x \cos^4 x \, dx \]

### Step 6: Change of variables
Next, we will use the substitution \( t = \cos x \), which gives \( dt = -\sin x \, dx \). The limits change as follows: when \( x = 0 \), \( t = 1 \) and when \( x = \pi \), \( t = -1 \).

So, we have:

\[ \int_{0}^{\pi} \sin x \cos^4 x \, dx = -\int_{1}^{-1} t^4 \, dt = \int_{-1}^{1} t^4 \, dt \]

### Step 7: Evaluate the integral
Now we compute the integral:

\[ \int_{-1}^{1} t^4 \, dt = 2 \int_{0}^{1} t^4 \, dt = 2 \left[ \frac{t^5}{5} \right]_{0}^{1} = 2 \cdot \frac{1}{5} = \frac{2}{5} \]

### Step 8: Substitute back to find \( I \)
Now substituting back into our expression for \( I \):

\[ I = \frac{\pi}{2} \cdot \frac{2}{5} = \frac{\pi}{5} \]

### Final Answer
Thus, the final result is:

\[ I = \frac{\pi}{5} \]