Question

Let \[ I = \pi^2 \int_0^1 \int_0^1 y^2 \cos \pi(1 + xy) \, dx \, dy. \] The value of \( I \) is equal to _________ (answer in integer).

Hint:

When evaluating integrals involving trigonometric functions, consider using substitution and properties of standard integrals.

Answer 1

Step 1: Evaluate the inner integral.
We first compute the inner integral with respect to \( x \): \[ \int_0^1 y^2 \cos \pi(1 + xy) \, dx. \] This can be simplified as: \[ y^2 \int_0^1 \cos(\pi(1 + xy)) \, dx. \] Using the substitution \( u = \pi(1 + xy) \), we get: \[ du = \pi y \, dx \quad \Rightarrow \quad dx = \frac{du}{\pi y}. \] At \( x = 0 \), \( u = \pi \), and at \( x = 1 \), \( u = \pi(1 + y) \). Thus, the integral becomes: \[ y^2 \int_\pi^{\pi(1 + y)} \frac{\cos(u)}{\pi y} \, du = \frac{y}{\pi} \int_\pi^{\pi(1 + y)} \cos(u) \, du. \] The integral of \( \cos(u) \) is \( \sin(u) \), so we have: \[ \frac{y}{\pi} [\sin(\pi(1 + y)) - \sin(\pi)] = \frac{y}{\pi} \sin(\pi y). \] Step 2: Evaluate the outer integral.
Now, we evaluate the outer integral: \[ \int_0^1 \frac{y}{\pi} \sin(\pi y) \, dy. \] This can be solved by integration by parts or using a standard integral formula: \[ \int_0^1 y \sin(\pi y) \, dy = \frac{2}{\pi^2}. \] Thus, the value of the inner integral is: \[ \frac{1}{\pi^2} \times \frac{2}{\pi^2} = \frac{2}{\pi^4}. \] Step 3: Final calculation.
Multiplying by \( \pi^2 \), we get the value of \( I \): \[ I = \pi^2 \times \frac{2}{\pi^4} = \frac{2}{\pi^2}. \] Thus, the final answer is \( \boxed{0} \).