The value of `int_(0)^(pi)|cos x|^(3)dx` is :

Options

A. 0
B. `-(4)/(3)`
C. `(4)/(3)`
D. `(2)/(3)`

Answer

To solve the integral \( I = \int_{0}^{\pi} |\cos x|^3 \, dx \), we will break it down into manageable steps.

### Step 1: Analyze the behavior of \(\cos x\)
The function \(\cos x\) is positive in the interval \([0, \frac{\pi}{2}]\) and negative in the interval \([\frac{\pi}{2}, \pi]\). Therefore, we can express the absolute value of \(\cos x\) as:
\[ |\cos x| = \begin{cases} \cos x & \text{for } 0 \leq x \leq \frac{\pi}{2} \\ -\cos x & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

### Step 2: Set up the integral
We can split the integral into two parts:
\[ I = \int_{0}^{\pi} |\cos x|^3 \, dx = \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x)^3 \, dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx - \int_{\frac{\pi}{2}}^{\pi} \cos^3 x \, dx \]

### Step 3: Change of variable for the second integral
For the second integral, we can change the variable. Let \( t = \pi - x \). Then \( dt = -dx \), and the limits change as follows:
- When \( x = \frac{\pi}{2} \), \( t = \frac{\pi}{2} \)
- When \( x = \pi \), \( t = 0 \)

Thus, we have:
\[ \int_{\frac{\pi}{2}}^{\pi} \cos^3 x \, dx = \int_{\frac{\pi}{2}}^{0} \cos^3(\pi - t)(-dt) = \int_{0}^{\frac{\pi}{2}} (-\cos t)^3 \, dt = -\int_{0}^{\frac{\pi}{2}} \cos^3 t \, dt \]

### Step 4: Combine the integrals
Now substituting back into the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx - (-\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx) = 2 \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx \]

### Step 5: Evaluate \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx\)
To evaluate \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx\), we can use the reduction formula:
\[ \int \cos^n x \, dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int \cos^{n-2} x \, dx \]
For \( n = 3 \):
\[ \int \cos^3 x \, dx = \frac{1}{3} \cos^2 x \sin x + \frac{2}{3} \int \cos x \, dx \]
Now, we know:
\[ \int \cos x \, dx = \sin x \]
Thus:
\[ \int \cos^3 x \, dx = \frac{1}{3} \cos^2 x \sin x + \frac{2}{3} \sin x \]
Evaluating from \(0\) to \(\frac{\pi}{2}\):
\[ \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = \left[ \frac{1}{3} \cos^2 x \sin x + \frac{2}{3} \sin x \right]_{0}^{\frac{\pi}{2}} = \left[ 0 + \frac{2}{3} \cdot 1 \right] - \left[ \frac{1}{3} \cdot 1 \cdot 0 + 0 \right] = \frac{2}{3} \]

### Step 6: Final Calculation
Substituting back into our expression for \(I\):
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = 2 \cdot \frac{2}{3} = \frac{4}{3} \]

Thus, the value of the integral \( \int_{0}^{\pi} |\cos x|^3 \, dx \) is:
\[ \boxed{\frac{4}{3}} \]