`int_(0)^(pi/2)sin^(2)x cos^(4)x dx=`

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(8)`
D. `(pi)/(32)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^4 x \, dx \), we can use a property of definite integrals and some trigonometric identities. Let's go through the solution step by step.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^4 x \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals:
\[ \int_{0}^{A} f(x) \, dx = \int_{0}^{A} f(A - x) \, dx \]
In our case, \( A = \frac{\pi}{2} \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^2\left(\frac{\pi}{2} - x\right) \cos^4\left(\frac{\pi}{2} - x\right) \, dx \]
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^2 x \sin^4 x \, dx \]

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

Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \sin^2 x \cos^4 x + \cos^2 x \sin^4 x \right) \, dx \]

### Step 4: Simplify the integrand
We can factor the integrand:
\[ \sin^2 x \cos^4 x + \cos^2 x \sin^4 x = \sin^2 x \cos^2 x (\cos^2 x + \sin^2 x) = \sin^2 x \cos^2 x \]
since \( \cos^2 x + \sin^2 x = 1 \).

Thus, we have:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx \]

### Step 5: Use the identity for \( \sin^2 x \cos^2 x \)
We can use the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \sin^2(2x) \, dx \]
This simplifies to:
\[ 2I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin^2(2x) \, dx \]

### Step 6: Evaluate the integral
The integral \( \int \sin^2(kx) \, dx \) can be evaluated using the formula:
\[ \int \sin^2(kx) \, dx = \frac{x}{2} - \frac{\sin(2kx)}{4k} + C \]
For our case with \( k = 2 \):
\[ \int \sin^2(2x) \, dx = \frac{x}{2} - \frac{\sin(4x)}{8} + C \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\frac{\pi}{2}} \sin^2(2x) \, dx = \left[ \frac{x}{2} - \frac{\sin(4x)}{8} \right]_{0}^{\frac{\pi}{2}} = \left[ \frac{\pi/4} - 0 \right] - \left[ 0 - 0 \right] = \frac{\pi}{4} \]

### Step 7: Substitute back to find \( I \)
Now substituting back:
\[ 2I = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16} \]
Thus,
\[ I = \frac{\pi}{32} \]

### Final Answer
The value of the integral is:
\[ I = \frac{\pi}{32} \]