`int_(0)^((pi)/(2))(sin^(3)x)/(sinx+cosx)dx` is equal to

Options

A. (a) `(pi-2)/(8)`
B. (b) `(pi-2)/(4)`
C. (c) `(pi-1)/(4)`
D. (d) `(pi-1)/(2)`

Answer

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

### Step 1: Define the integral
Let:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx \]

### Step 2: Use the property of integrals
Using the property that \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \) for \( a = \frac{\pi}{2} \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx \]

### Step 3: Simplify using trigonometric identities
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos x + \sin x} \, dx \]

### Step 4: Add the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx \) (Equation 1)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx \) (Equation 2)

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

### Step 5: Use the identity for \( a^3 + b^3 \)
Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x) \]

### Step 6: Substitute back into the integral
Substituting this back into our equation for \( 2I \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x + \cos x)(1 - \sin x \cos x)}{\sin x + \cos x} \, dx \]
This simplifies to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} (1 - \sin x \cos x) \, dx \]

### Step 7: Split the integral
Now we can split this integral:
\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx \]

### Step 8: Evaluate the integrals
The first integral:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]
The second integral can be evaluated using the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\[ \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx = \frac{1}{2} \left[-\frac{1}{2} \cos 2x\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left[0 - (-\frac{1}{2})\right] = \frac{1}{4} \]

### Step 9: Substitute back to find \( I \)
Now substituting back:
\[ 2I = \frac{\pi}{2} - \frac{1}{4} \]
Thus,
\[ 2I = \frac{\pi}{2} - \frac{1}{4} = \frac{2\pi - 1}{4} \]
Dividing both sides by 2:
\[ I = \frac{2\pi - 1}{8} \]

### Final Result
Thus, the final value of the integral is:
\[ I = \frac{\pi - \frac{1}{2}}{4} = \frac{\pi - 1}{4} \]

### Conclusion
The value of the integral \( \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx \) is:
\[ \boxed{\frac{\pi - 1}{4}} \]