The value of `underset(0)overset(pi//2)int (sin^(3)x)/(sin x + cos x)dx` is `(pi-1)/(k)` . The value of `k` is ________.

Explanation

To solve the integral

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx, \]

we will use a symmetry property of the integral and some algebraic manipulation.

### Step 1: Write the integral

We start with the integral:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx. \]

### Step 2: Use the substitution \( x = \frac{\pi}{2} - t \)

Using the substitution \( x = \frac{\pi}{2} - t \), we have \( dx = -dt \). The limits change as follows: when \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \). Thus, we can rewrite the integral:

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

This simplifies to:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 t}{\cos t + \sin t} \, dt. \]

### Step 3: Combine the two integrals

Now we have two expressions for \( I \):

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx, \]
\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx. \]

Adding these two integrals gives:

\[ 2I = \int_0^{\frac{\pi}{2}} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} \, dx. \]

### Step 4: Simplify the numerator using the identity

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 \( \sin^2 x + \cos^2 x = 1 \), we can write:

\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x). \]

### Step 5: Substitute back into the integral

Substituting this back into our equation for \( 2I \):

\[ 2I = \int_0^{\frac{\pi}{2}} (1 - \sin x \cos x) \, dx. \]

### Step 6: Evaluate the integral

Now we can 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 \( \sin x \cos x = \frac{1}{2} \sin(2x) \):

\[ \int_0^{\frac{\pi}{2}} \sin x \cos x \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) \, dx = \frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left(0 - (-\frac{1}{2})\right) = \frac{1}{4}. \]

### Step 7: Combine results

Thus,

\[ 2I = \frac{\pi}{2} - \frac{1}{4}. \]

### Step 8: Solve for \( I \)

Now, we can solve for \( I \):

\[ 2I = \frac{\pi}{2} - \frac{1}{4} = \frac{2\pi - 1}{4} \implies I = \frac{2\pi - 1}{8}. \]

### Step 9: Compare with the given expression

Given that:

\[ I = \frac{\pi - 1}{k}, \]

we can set:

\[ \frac{2\pi - 1}{8} = \frac{\pi - 1}{k}. \]

Cross-multiplying gives:

\[ k(2\pi - 1) = 8(\pi - 1). \]

### Step 10: Solve for \( k \)

Expanding and rearranging:

\[ 2k\pi - k = 8\pi - 8 \implies 2k\pi - 8\pi = k - 8 \implies (2k - 8)\pi = k - 8. \]

Setting coefficients equal, we find:

\[ 2k - 8 = 0 \implies 2k = 8 \implies k = 4. \]

Thus, the value of \( k \) is

\[ \boxed{4}. \]