The problem consists of two parts. First, we are asked to evaluate the integral $\int_0^{\pi/2} x^2 \cos(2x) \, dx$ using integration by parts twice. Second, we are given $I = \int_0^{\pi/2} x^2 \cos^2(x) \, dx$ and $J = \int_0^{\pi/2} x^2 \sin^2(x) \, dx$. We are asked to calculate $I+J$ and $I-J$ and then find $I$ and $J$.

1. Problem Description

The problem consists of two parts.

First, we are asked to evaluate the integral $\int_0^{\pi/2} x^2 \cos(2x) \, dx$ using integration by parts twice.

Second, we are given $I = \int_0^{\pi/2} x^2 \cos^2(x) \, dx$ and $J = \int_0^{\pi/2} x^2 \sin^2(x) \, dx$. We are asked to calculate $I+J$ and $I-J$ and then find $I$ and $J$.

2. Solution Steps

Part 1: Evaluate $\int_0^{\pi/2} x^2 \cos(2x) \, dx$.

We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = x^2$ and $dv = \cos(2x) \, dx$. Then $du = 2x \, dx$ and $v = \frac{1}{2} \sin(2x)$.

So,

$\int_0^{\pi/2} x^2 \cos(2x) \, dx = \left[ x^2 \cdot \frac{1}{2} \sin(2x) \right]_0^{\pi/2} - \int_0^{\pi/2} \frac{1}{2} \sin(2x) \cdot 2x \, dx$

$= \left[ \frac{1}{2} x^2 \sin(2x) \right]_0^{\pi/2} - \int_0^{\pi/2} x \sin(2x) \, dx$

$= \frac{1}{2} \left( (\frac{\pi}{2})^2 \sin(\pi) - 0^2 \sin(0) \right) - \int_0^{\pi/2} x \sin(2x) \, dx$

$= 0 - \int_0^{\pi/2} x \sin(2x) \, dx$

$= - \int_0^{\pi/2} x \sin(2x) \, dx$

We use integration by parts again.

Let $u = x$ and $dv = \sin(2x) \, dx$. Then $du = dx$ and $v = -\frac{1}{2} \cos(2x)$.

So,

$-\int_0^{\pi/2} x \sin(2x) \, dx = - \left( \left[ x \cdot (-\frac{1}{2} \cos(2x)) \right]_0^{\pi/2} - \int_0^{\pi/2} (-\frac{1}{2} \cos(2x)) \, dx \right)$

$= - \left( \left[ -\frac{1}{2} x \cos(2x) \right]_0^{\pi/2} + \frac{1}{2} \int_0^{\pi/2} \cos(2x) \, dx \right)$

$= - \left( -\frac{1}{2} \frac{\pi}{2} \cos(\pi) - (-\frac{1}{2} \cdot 0 \cdot \cos(0)) + \frac{1}{2} \left[ \frac{1}{2} \sin(2x) \right]_0^{\pi/2} \right)$

$= - \left( -\frac{\pi}{4} (-1) - 0 + \frac{1}{4} (\sin(\pi) - \sin(0)) \right)$

$= - \left( \frac{\pi}{4} + \frac{1}{4} (0 - 0) \right)$

$= -\frac{\pi}{4}$

Part 2: Evaluate $I+J$ and $I-J$ and find $I$ and $J$.

$I = \int_0^{\pi/2} x^2 \cos^2(x) \, dx$

$J = \int_0^{\pi/2} x^2 \sin^2(x) \, dx$

$I+J = \int_0^{\pi/2} x^2 (\cos^2(x) + \sin^2(x)) \, dx = \int_0^{\pi/2} x^2 \, dx = \left[ \frac{1}{3} x^3 \right]_0^{\pi/2} = \frac{1}{3} (\frac{\pi}{2})^3 - 0 = \frac{\pi^3}{24}$

$I-J = \int_0^{\pi/2} x^2 (\cos^2(x) - \sin^2(x)) \, dx = \int_0^{\pi/2} x^2 \cos(2x) \, dx = -\frac{\pi}{4}$ (from Part 1)

Now we have:

$I+J = \frac{\pi^3}{24}$

$I-J = -\frac{\pi}{4}$

Adding the two equations:

$2I = \frac{\pi^3}{24} - \frac{\pi}{4} = \frac{\pi^3 - 6\pi}{24}$

$I = \frac{\pi^3 - 6\pi}{48}$

Subtracting the two equations:

$2J = \frac{\pi^3}{24} + \frac{\pi}{4} = \frac{\pi^3 + 6\pi}{24}$

$J = \frac{\pi^3 + 6\pi}{48}$

3. Final Answer

$\int_0^{\pi/2} x^2 \cos(2x) \, dx = -\frac{\pi}{4}$

$I+J = \frac{\pi^3}{24}$

$I-J = -\frac{\pi}{4}$

$I = \frac{\pi^3 - 6\pi}{48}$

$J = \frac{\pi^3 + 6\pi}{48}$