The value of `int_(0)^((pi)/(2))(cos 2x cos 2^(2)x cos 2^(3)x cos 2^(4)x)dx` is equal to

Options

A. 0
B. `(1)/(2)`
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \cos(2x) \cos(2^2 x) \cos(2^3 x) \cos(2^4 x) \, dx, \]

we can use the property of definite integrals that states:

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

In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \), so we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \cos(2(\frac{\pi}{2} - x)) \cos(2^2(\frac{\pi}{2} - x)) \cos(2^3(\frac{\pi}{2} - x)) \cos(2^4(\frac{\pi}{2} - x)) \, dx. \]

Now, we compute each cosine term:

1. For \( \cos(2(\frac{\pi}{2} - x)) = \cos(\pi - 2x) = -\cos(2x) \).
2. For \( \cos(2^2(\frac{\pi}{2} - x)) = \cos(2^2(\frac{\pi}{2} - x)) = \cos(2^2 \cdot \frac{\pi}{2} - 2^2 x) = \cos(2^2 \cdot \frac{\pi}{2} - 4x) = \cos(2^2 \cdot \frac{\pi}{2}) \cos(4x) + \sin(2^2 \cdot \frac{\pi}{2}) \sin(4x) = 0 \cdot \cos(4x) + 1 \cdot \sin(4x) = \sin(4x) \).
3. For \( \cos(2^3(\frac{\pi}{2} - x)) = \cos(2^3 \cdot \frac{\pi}{2} - 8x) = \cos(2^3 \cdot \frac{\pi}{2}) \cos(8x) + \sin(2^3 \cdot \frac{\pi}{2}) \sin(8x) = 0 \cdot \cos(8x) + 1 \cdot \sin(8x) = \sin(8x) \).
4. For \( \cos(2^4(\frac{\pi}{2} - x)) = \cos(2^4 \cdot \frac{\pi}{2} - 16x) = \cos(2^4 \cdot \frac{\pi}{2}) \cos(16x) + \sin(2^4 \cdot \frac{\pi}{2}) \sin(16x) = 0 \cdot \cos(16x) + 1 \cdot \sin(16x) = \sin(16x) \).

Putting it all together, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} (-\cos(2x)) \sin(4x) \sin(8x) \sin(16x) \, dx. \]

Now we can add the two expressions for \( I \):

\[ 2I = \int_{0}^{\frac{\pi}{2}} \cos(2x) \cos(2^2 x) \cos(2^3 x) \cos(2^4 x) \, dx + \int_{0}^{\frac{\pi}{2}} (-\cos(2x)) \sin(4x) \sin(8x) \sin(16x) \, dx. \]

Since the two integrals are equal in magnitude but opposite in sign, we have:

\[ 2I = 0 \implies I = 0. \]

Thus, the value of the integral is

\[ \boxed{0}. \]