The problem asks us to evaluate the integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

1. Problem Description

The problem asks us to evaluate the integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

2. Solution Steps

Let $u = x \sin x + 9 \cos x$. Then,

$$du = (\sin x + x \cos x - 9 \sin x) dx = (x \cos x - 8 \sin x) dx$$

We want to express $x^2 + 72$ in terms of $u$ and $du$. Notice that

$$\frac{x \cos x - 8 \sin x}{x \sin x + 9 \cos x}$$

doesn't seem to help. Instead, consider

$$\frac{d}{dx} \left( \frac{f(x)}{x \sin x + 9 \cos x} \right) = \frac{f'(x) (x \sin x + 9 \cos x) - f(x) (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$$

Let $f(x) = A x + B$, so $f'(x) = A$. Then

$$\frac{A (x \sin x + 9 \cos x) - (Ax + B) (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$$

$$= \frac{Ax \sin x + 9A \cos x - Ax^2 \cos x + 8Ax \sin x - Bx \cos x + 8B \sin x}{(x \sin x + 9 \cos x)^2}$$

$$= \frac{-Ax^2 \cos x + (9A-Bx) \cos x + (9Ax+8B) \sin x }{(x \sin x + 9 \cos x)^2}$$

Consider $\frac{x}{x \sin x + 9 \cos x}$. The derivative is

$$\frac{(x \sin x + 9 \cos x) - x (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{x \sin x + 9 \cos x - x^2 \cos x + 8x \sin x}{(x \sin x + 9 \cos x)^2} = \frac{-x^2 \cos x + 9x \sin x + 9 \cos x}{(x \sin x + 9 \cos x)^2}$$

Consider $\frac{\sin x}{x \sin x + 9 \cos x}$. The derivative is

$$\frac{\cos x (x \sin x + 9 \cos x) - \sin x (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{x \sin x \cos x + 9 \cos^2 x - x \sin x \cos x + 8 \sin^2 x}{(x \sin x + 9 \cos x)^2} = \frac{9 \cos^2 x + 8 \sin^2 x}{(x \sin x + 9 \cos x)^2}$$

Consider $\frac{\cos x}{x \sin x + 9 \cos x}$. The derivative is

$$\frac{-\sin x (x \sin x + 9 \cos x) - \cos x (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{-x \sin^2 x - 9 \sin x \cos x - x \cos^2 x + 8 \sin x \cos x}{(x \sin x + 9 \cos x)^2} = \frac{-x - \sin x \cos x}{(x \sin x + 9 \cos x)^2}$$

Try the derivative of $\frac{Ax+B}{x \sin x + 9 \cos x}$:

$$\frac{A(x \sin x + 9 \cos x) - (Ax+B)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{Ax \sin x + 9A \cos x - Ax^2 \cos x + 8Ax \sin x - Bx \cos x + 8B \sin x}{(x \sin x + 9 \cos x)^2} = \frac{-Ax^2 \cos x + (9A-Bx) \cos x + (9A+8B)x \sin x + 8B \sin x}{(x \sin x + 9 \cos x)^2}$$

Try something simpler.

Notice that

$\frac{d}{dx} \frac{\sin x}{x \sin x + 9 \cos x} = \frac{(x \sin x + 9 \cos x)\cos x - (\sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{x \sin x \cos x + 9 \cos^2 x - x \sin x \cos x + 8 \sin^2 x}{(x \sin x + 9 \cos x)^2} = \frac{9 \cos^2 x + 8 \sin^2 x}{(x \sin x + 9 \cos x)^2}$

$\frac{d}{dx} \frac{x \cos x}{x \sin x + 9 \cos x} = \frac{(x \sin x + 9 \cos x)(\cos x - x \sin x) - (x \cos x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{x \sin x \cos x + 9 \cos^2 x - x^2 \sin^2 x - 9x \sin x \cos x - x^2 \cos^2 x + 8 x \cos x \sin x}{(x \sin x + 9 \cos x)^2} = \frac{9 \cos^2 x - x^2 - 8 x \sin x \cos x}{(x \sin x + 9 \cos x)^2}$

Let's look at $\frac{x \cos x - 9 \sin x}{x \sin x + 9 \cos x}$. Taking the derivative yields $\frac{(x \sin x + 9 \cos x)(\cos x - x \sin x - 9 \cos x) - (x \cos x - 9 \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{(x \sin x + 9 \cos x)(-x \sin x - 8 \cos x) - (x \cos x - 9 \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = -\frac{x^2+72}{(x \sin x + 9 \cos x)^2}$. Therefore, $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx = -\frac{x \cos x - 9 \sin x}{x \sin x + 9 \cos x} + C$.

3. Final Answer

$-\frac{x \cos x - 9 \sin x}{x \sin x + 9 \cos x} + C$