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

Options

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

Answer

To solve the integral

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

we can use the property of definite integrals. The property states that:

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

In this case, we will set \( a = \frac{\pi}{2} \).

### Step 1: Apply the property

Let’s compute \( I \) using the property:

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

Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos x + \sin x} \, dx. \]

### Step 2: Combine the integrals

Now we can add both expressions for \( I \):

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sin^3 x}{\sin x + \cos x} + \frac{\cos^3 x}{\sin x + \cos x} \right) \, dx. \]

This simplifies to:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} \, dx. \]

### Step 3: Simplify the numerator

Using the identity for the sum of cubes, we have:

\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x). \]

Since \( \sin^2 x + \cos^2 x = 1 \), we can rewrite this as:

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

### Step 4: Substitute back into the integral

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

\[ 2I = \int_{0}^{\frac{\pi}{2}} (1 - \sin x \cos x) \, dx. \]

### Step 5: Evaluate the integral

Now we can evaluate the integral:

\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx. \]

The first integral is straightforward:

\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}. \]

For the second integral, we use the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \):

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

### Step 6: Combine results

Now we can combine the results:

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

### Step 7: Solve for \( I \)

Thus,

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

Dividing by 2 gives:

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

### Final Answer

Therefore, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx = \frac{2\pi - 1}{8}. \]