`int_(0)^((pi)/2)(cos^(3)x)/(sinx+cosx)dx`

Options

A. `(pi-1)/2`
B. `(pi-1)/4`
C. `(1+pi)/4`
D. `(pi-3)/4`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx \), we can use the property of definite integrals.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx \]

2. **Use the Property of Definite Integrals**:
We can use the substitution \( x = \frac{\pi}{2} - t \). Then, \( dx = -dt \) and the limits change as follows:
- When \( x = 0 \), \( t = \frac{\pi}{2} \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)

Thus, we have:
\[ I = \int_{\frac{\pi}{2}}^{0} \frac{\cos^3\left(\frac{\pi}{2} - t\right)}{\sin\left(\frac{\pi}{2} - t\right) + \cos\left(\frac{\pi}{2} - t\right)} (-dt) \]

3. **Simplify the Integral**:
Using the identities \( \cos\left(\frac{\pi}{2} - t\right) = \sin t \) and \( \sin\left(\frac{\pi}{2} - t\right) = \cos t \), we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 t}{\cos t + \sin t} \, dt \]

4. **Combine the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx \]

Adding these two integrals gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x + \sin^3 x}{\sin x + \cos x} \, dx \]

5. **Use the Identity for Sums of Cubes**:
Recall the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\[ \cos^3 x + \sin^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]

Since \( \sin^2 x + \cos^2 x = 1 \):
\[ \cos^3 x + \sin^3 x = (\sin x + \cos x)(1 - \sin x \cos x) \]

6. **Substitute Back into the Integral**:
Substitute this back into the integral:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x + \cos x)(1 - \sin x \cos x)}{\sin x + \cos x} \, dx \]
This simplifies to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} (1 - \sin x \cos x) \, dx \]

7. **Evaluate the Integral**:
Split the integral:
\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx \]
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]
The second integral can be evaluated using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):
\[ \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx = \frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

8. **Combine Results**:
Thus,
\[ 2I = \frac{\pi}{2} - \frac{1}{4} \]
\[ 2I = \frac{2\pi}{4} - \frac{1}{4} = \frac{2\pi - 1}{4} \]
Therefore,
\[ I = \frac{2\pi - 1}{8} \]

### Final Answer:
\[ I = \frac{2\pi - 1}{8} \]