If `int_(0)^(pi//2)sin^(4)x cos^(2)x dx=(pi)/(32)`, then `int_(0)^(pi//2)sin^(2)x cos^(4)x dx=`

Options

A. `(pi)/(4)`
B. `(pi)/(8)`
C. `(pi)/(16)`
D. `(pi)/(32)`

Answer

To solve the problem, we need to find the value of the integral

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

We are given that

\[ \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^2 x \, dx = \frac{\pi}{32}. \]

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

We can use the property of definite integrals which states that:

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

In our case, let \( a = \frac{\pi}{2} \). Then we can write:

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

### Step 2: Simplify the integrals

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:

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

### Step 3: Relate the two integrals

Now we have:

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

Let’s denote:

\[ I_1 = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^2 x \, dx, \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^2 x \, dx. \]

From the above, we can conclude that:

\[ I_1 = I_2. \]

### Step 4: Find the total integral

Now, we can express the total integral:

\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^2 x \, dx + \int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^2 x \, dx. \]

This can be simplified as:

\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx. \]

### Step 5: Evaluate the integral

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ I_1 + I_2 = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx. \]

Now, we know:

\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin^2(2x) \, dx = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16}. \]

### Step 6: Solve for \( I \)

Since \( I_1 = I_2 \):

\[ 2I_1 = \frac{\pi}{16} \implies I_1 = \frac{\pi}{32}. \]

Thus, we have:

\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^4 x \, dx = \frac{\pi}{32}. \]

### Final Answer

Therefore, the value of

\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^4 x \, dx = \frac{\pi}{32}. \]