We want to evaluate the integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

1. Problem Description

We want to evaluate the integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

2. Solution Steps

Let's try to find a function whose derivative is close to the integrand. We note that

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

This doesn't seem to help much.

Instead, consider $\frac{d}{dx} \left(\frac{f(x)}{x \sin x + 9 \cos x}\right)$ for some function $f(x)$.

Using the quotient rule, we have

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

If we choose $f(x) = x \cos x - 8 \sin x$, then $f'(x) = \cos x - x \sin x - 8 \cos x = -x \sin x - 7 \cos x$.

Then

$\frac{d}{dx} \left(\frac{x \cos x - 8 \sin x}{x \sin x + 9 \cos x}\right) = \frac{(-x \sin x - 7 \cos x)(x \sin x + 9 \cos x) - (x \cos x - 8 \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$

$= \frac{-x^2 \sin^2 x - 9x \sin x \cos x - 7x \sin x \cos x - 63 \cos^2 x - (x^2 \cos^2 x - 8x \sin x \cos x - 8x \sin x \cos x + 64 \sin^2 x)}{(x \sin x + 9 \cos x)^2}$

$= \frac{-x^2 \sin^2 x - 16x \sin x \cos x - 63 \cos^2 x - x^2 \cos^2 x + 16x \sin x \cos x - 64 \sin^2 x}{(x \sin x + 9 \cos x)^2}$

$= \frac{-x^2 (\sin^2 x + \cos^2 x) - (63 \cos^2 x + 64 \sin^2 x)}{(x \sin x + 9 \cos x)^2} = \frac{-x^2 - 63 \cos^2 x - 64 \sin^2 x}{(x \sin x + 9 \cos x)^2}$

$= \frac{-x^2 - 63 (\cos^2 x + \sin^2 x) - \sin^2 x}{(x \sin x + 9 \cos x)^2} = \frac{-x^2 - 63 - \sin^2 x}{(x \sin x + 9 \cos x)^2}$

This doesn't work.

Consider $\frac{d}{dx} \frac{x \cos x + a \sin x}{x \sin x + 9 \cos x}$.

The derivative is $\frac{(\cos x - x \sin x + a \cos x)(x \sin x + 9 \cos x) - (x \cos x + a \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$

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

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

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

We want $9 + 9a = 72/9 = 8$ and $8a = 0$.

$9(1+a) = 8$ so $1+a = 8/9$ and $a = -1/9$. This doesn't work.

We can rewrite $x^2 + 72 = (x^2+81) - 9$.

Let's try $\frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x}$. Its derivative is

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

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

$= \frac{8x \sin x \cos x + 72 \cos^2 x + x^2 \sin^2 x + 9x \sin x \cos x - (9x \sin x \cos x - 72 \sin^2 x - x^2 \cos^2 x + 8x \sin x \cos x)}{(x \sin x + 9 \cos x)^2}$

$= \frac{8x \sin x \cos x + 72 \cos^2 x + x^2 \sin^2 x + 9x \sin x \cos x - 9x \sin x \cos x + 72 \sin^2 x + x^2 \cos^2 x - 8x \sin x \cos x}{(x \sin x + 9 \cos x)^2}$

$= \frac{x^2(\sin^2 x + \cos^2 x) + 72(\cos^2 x + \sin^2 x)}{(x \sin x + 9 \cos x)^2} = \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2}$.

Thus, $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx = \frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x} + C$.

3. Final Answer

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