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 can use the property of definite integrals and some algebraic manipulations. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^6 x \, dx \]

### Step 2: Use the Symmetry Property of Integrals
Using the property of integrals, we know that:
\[ \int_{0}^{A} f(x) \, dx = \int_{0}^{A} f(A - x) \, dx \]
For our case, we can write:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^6\left(\frac{\pi}{2} - x\right) \, dx \]
Since \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^6 x \, dx \]

### Step 3: Add the Two Integrals
Now we can add the two 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 Sine and Cosine
We can use the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) where \( a = \sin^2 x \) 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 Further
Now, we can express \( \sin^4 x + \cos^4 x \) using the identity:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus,
\[ \sin^6 x + \cos^6 x = (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x \]

### 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 evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]

### Step 8: Evaluate the Second Integral
For the second integral, we can 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 \( \int \sin^2 kx \, dx = \frac{x}{2} - \frac{\sin 2kx}{4k} + C \):
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 2x \, dx = \frac{\pi}{4} \]
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
Now substituting back:
\[ 2I = \frac{\pi}{2} - 3 \cdot \frac{\pi}{16} \]
Calculating this gives:
\[ 2I = \frac{\pi}{2} - \frac{3\pi}{16} = \frac{8\pi}{16} - \frac{3\pi}{16} = \frac{5\pi}{16} \]
Thus,
\[ I = \frac{5\pi}{32} \]

### Final Answer
The value of the integral \( \int_{0}^{\frac{\pi}{2}} \sin^6 x \, dx \) is:
\[ \boxed{\frac{5\pi}{32}} \]