The value of `int_(0)^(pi//2) (sin^(3)x cos x)/(sin^(4)x+ cos^(4)x )dx` is

Options

A. `pi//8`
B. `pi//4`
C. `pi//2`
D. `pi`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x \cos x}{\sin^4 x + \cos^4 x} \, dx, \]

we can use the property of definite integrals that states:

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

### Step 1: Apply the property of definite integrals

Let’s first apply this property to our integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3\left(\frac{\pi}{2} - x\right) \cos\left(\frac{\pi}{2} - x\right)}{\sin^4\left(\frac{\pi}{2} - x\right) + \cos^4\left(\frac{\pi}{2} - x\right)} \, dx. \]

### Step 2: Simplify the integrand

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

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x \sin x}{\cos^4 x + \sin^4 x} \, dx. \]

### Step 3: Combine the two integrals

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

1. \(I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x \cos x}{\sin^4 x + \cos^4 x} \, dx\)
2. \(I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x \sin x}{\sin^4 x + \cos^4 x} \, dx\)

Adding these two expressions gives:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x \cos x + \cos^3 x \sin x}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 4: Factor the numerator

Notice that the numerator can be factored:

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

Thus, we have:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx. \]

### Step 5: Simplify the denominator

We can simplify the denominator:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x. \]

### Step 6: Substitute and evaluate

Now we substitute back into the integral:

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

Using the identity \(\sin 2x = 2 \sin x \cos x\), we can rewrite the integral:

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

### Step 7: Solve the integral

This integral can be evaluated using standard techniques or tables, leading to:

\[ I = \frac{\pi}{8}. \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{8}}. \]