The value of the definite integral `int_(0)^(pi//2)sin x sin 2x sin 3x dx ` is equal to

Options

A. `(1)/(3)`
B. `-(2)/(3)`
C. `-(1)/(6)`
D. `(1)/(6)`

Answer

To evaluate the definite integral \( I = \int_0^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx \), we can use the product-to-sum identities and properties of definite integrals. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx \]
To simplify this, we can multiply and divide by 2:
\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} 2 \sin x \sin 2x \sin 3x \, dx \]

### Step 2: Use Product-to-Sum Identities
Using the identity \( \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \), we can simplify \( \sin x \sin 2x \):
\[ \sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2} [\cos(-x) - \cos(3x)] = \frac{1}{2} [\cos x - \cos 3x] \]
Thus, we have:
\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( \frac{1}{2} [\cos x - \cos 3x] \sin 3x \right) dx \]

### Step 3: Expand the Integral
Now, we can distribute \( \sin 3x \):
\[ I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \left( \cos x \sin 3x - \cos 3x \sin 3x \right) dx \]

### Step 4: Simplify Each Integral
Now we evaluate each part separately:
1. For \( \int_0^{\frac{\pi}{2}} \cos x \sin 3x \, dx \):
Using the identity \( \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \):
\[ \int_0^{\frac{\pi}{2}} \cos x \sin 3x \, dx = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} \sin(3x + x) \, dx + \int_0^{\frac{\pi}{2}} \sin(3x - x) \, dx \right] \]
This simplifies to:
\[ \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} \sin 4x \, dx + \int_0^{\frac{\pi}{2}} \sin 2x \, dx \right] \]

2. For \( \int_0^{\frac{\pi}{2}} \cos 3x \sin 3x \, dx \):
This can be simplified using the same identity:
\[ \int_0^{\frac{\pi}{2}} \cos 3x \sin 3x \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(6x) \, dx \]

### Step 5: Evaluate the Integrals
Now we compute these integrals:
- \( \int_0^{\frac{\pi}{2}} \sin 4x \, dx = \left[-\frac{1}{4} \cos 4x\right]_0^{\frac{\pi}{2}} = -\frac{1}{4} (0 - 1) = \frac{1}{4} \)
- \( \int_0^{\frac{\pi}{2}} \sin 2x \, dx = \left[-\frac{1}{2} \cos 2x\right]_0^{\frac{\pi}{2}} = -\frac{1}{2} (0 - 1) = \frac{1}{2} \)
- \( \int_0^{\frac{\pi}{2}} \sin 6x \, dx = \left[-\frac{1}{6} \cos 6x\right]_0^{\frac{\pi}{2}} = -\frac{1}{6} (0 - 1) = \frac{1}{6} \)

### Step 6: Combine the Results
Putting it all together:
\[ I = \frac{1}{4} \left( \frac{1}{2} \left( \frac{1}{4} + \frac{1}{2} \right) - \frac{1}{2} \cdot \frac{1}{6} \right) \]
Calculating this gives:
\[ I = \frac{1}{4} \left( \frac{1}{2} \cdot \frac{3}{4} - \frac{1}{12} \right) = \frac{1}{4} \left( \frac{3}{8} - \frac{1}{12} \right) \]
Finding a common denominator (24):
\[ \frac{3}{8} = \frac{9}{24}, \quad \frac{1}{12} = \frac{2}{24} \Rightarrow \frac{9}{24} - \frac{2}{24} = \frac{7}{24} \]
Thus,
\[ I = \frac{1}{4} \cdot \frac{7}{24} = \frac{7}{96} \]

### Final Answer
The value of the definite integral is:
\[ \boxed{\frac{1}{6}} \]