The value of the integral `int_(0)^(pi//2) sin^(6) x dx`, is

Options

A. `(3pi)/(4)`
B. `(5)/(32) pi`
C. `(3)/(16) pi`
D. none of these

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \sin^6 x \, dx \), we will use the property of definite integrals and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let
\[ I = \int_0^{\frac{\pi}{2}} \sin^6 x \, dx \]

### 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, we can write:
\[ I = \int_0^{\frac{\pi}{2}} \sin^6 x \, dx = \int_0^{\frac{\pi}{2}} \cos^6 x \, dx \]

### Step 3: Add the two integrals
Now, we add both expressions for \( I \):
\[ 2I = \int_0^{\frac{\pi}{2}} \sin^6 x \, dx + \int_0^{\frac{\pi}{2}} \cos^6 x \, dx \]
This simplifies to:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \sin^6 x + \cos^6 x \right) \, dx \]

### Step 4: Use the identity for \( a^3 + b^3 \)
We can use the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \):
\[ \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^6 x + \cos^6 x = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x \]

### Step 5: Simplify \( \sin^4 x + \cos^4 x \)
Using the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \):
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]
Thus, we can write:
\[ \sin^6 x + \cos^6 x = (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 6: Substitute back into the integral
Now substituting back into our integral:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( 1 - 3\sin^2 x \cos^2 x \right) \, dx \]
This can be split into two integrals:
\[ 2I = \int_0^{\frac{\pi}{2}} 1 \, dx - 3 \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx \]

### Step 7: Evaluate the first integral
The first integral is straightforward:
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]

### Step 8: Evaluate the second integral
For the second integral, we use the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \):
\[ \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx = \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2(2x) \, dx \]
Using the formula \( \int \sin^2(kx) \, dx = \frac{x}{2} - \frac{\sin(2kx)}{4k} + C \), we find:
\[ \int_0^{\frac{\pi}{2}} \sin^2(2x) \, dx = \frac{\pi}{4} \]
Thus:
\[ \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16} \]

### Step 9: Substitute back to find \( I \)
Now substituting back:
\[ 2I = \frac{\pi}{2} - 3 \cdot \frac{\pi}{16} = \frac{\pi}{2} - \frac{3\pi}{16} = \frac{8\pi}{16} - \frac{3\pi}{16} = \frac{5\pi}{16} \]
Thus:
\[ I = \frac{5\pi}{32} \]

### Final Answer
The value of the integral \( \int_0^{\frac{\pi}{2}} \sin^6 x \, dx \) is
\[ \boxed{\frac{5\pi}{32}} \]