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

AnalysisIntegrationCalculusTrigonometryQuotient Rule
2025/3/11

1. Problem Description

We want to evaluate the integral ∫x2+72(xsin⁡x+9cos⁡x)2dx\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
ddx(xsin⁡x+9cos⁡x)=sin⁡x+xcos⁡x−9sin⁡x=xcos⁡x−8sin⁡x\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 ddx(f(x)xsin⁡x+9cos⁡x)\frac{d}{dx} \left(\frac{f(x)}{x \sin x + 9 \cos x}\right) for some function f(x)f(x).
Using the quotient rule, we have
ddx(f(x)xsin⁡x+9cos⁡x)=f′(x)(xsin⁡x+9cos⁡x)−f(x)(xcos⁡x−8sin⁡x)(xsin⁡x+9cos⁡x)2\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)=xcos⁡x−8sin⁡xf(x) = x \cos x - 8 \sin x, then f′(x)=cos⁡x−xsin⁡x−8cos⁡x=−xsin⁡x−7cos⁡xf'(x) = \cos x - x \sin x - 8 \cos x = -x \sin x - 7 \cos x.
Then
ddx(xcos⁡x−8sin⁡xxsin⁡x+9cos⁡x)=(−xsin⁡x−7cos⁡x)(xsin⁡x+9cos⁡x)−(xcos⁡x−8sin⁡x)(xcos⁡x−8sin⁡x)(xsin⁡x+9cos⁡x)2\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}
=−x2sin⁡2x−9xsin⁡xcos⁡x−7xsin⁡xcos⁡x−63cos⁡2x−(x2cos⁡2x−8xsin⁡xcos⁡x−8xsin⁡xcos⁡x+64sin⁡2x)(xsin⁡x+9cos⁡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}
=−x2sin⁡2x−16xsin⁡xcos⁡x−63cos⁡2x−x2cos⁡2x+16xsin⁡xcos⁡x−64sin⁡2x(xsin⁡x+9cos⁡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}
=−x2(sin⁡2x+cos⁡2x)−(63cos⁡2x+64sin⁡2x)(xsin⁡x+9cos⁡x)2=−x2−63cos⁡2x−64sin⁡2x(xsin⁡x+9cos⁡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}
=−x2−63(cos⁡2x+sin⁡2x)−sin⁡2x(xsin⁡x+9cos⁡x)2=−x2−63−sin⁡2x(xsin⁡x+9cos⁡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 ddxxcos⁡x+asin⁡xxsin⁡x+9cos⁡x\frac{d}{dx} \frac{x \cos x + a \sin x}{x \sin x + 9 \cos x}.
The derivative is (cos⁡x−xsin⁡x+acos⁡x)(xsin⁡x+9cos⁡x)−(xcos⁡x+asin⁡x)(xcos⁡x−8sin⁡x)(xsin⁡x+9cos⁡x)2\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}
=xsin⁡xcos⁡x+9cos⁡2x−x2sin⁡2x−9xsin⁡xcos⁡x+axsin⁡xcos⁡x+9acos⁡2x−x2cos⁡2x+8xsin⁡xcos⁡x−axsin⁡xcos⁡x+8asin⁡2x(xsin⁡x+9cos⁡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}
=−x2(sin⁡2x+cos⁡2x)+xsin⁡xcos⁡x−9xsin⁡xcos⁡x+8xsin⁡xcos⁡x+9cos⁡2x+9acos⁡2x+8asin⁡2x(xsin⁡x+9cos⁡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}
=−x2+(9+9a)cos⁡2x+8asin⁡2x(xsin⁡x+9cos⁡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=89 + 9a = 72/9 = 8 and 8a=08a = 0.
9(1+a)=89(1+a) = 8 so 1+a=8/91+a = 8/9 and a=−1/9a = -1/9. This doesn't work.
We can rewrite x2+72=(x2+81)−9x^2 + 72 = (x^2+81) - 9.
Let's try 9sin⁡x−xcos⁡xxsin⁡x+9cos⁡x\frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x}. Its derivative is
(9cos⁡x−cos⁡x+xsin⁡x)(xsin⁡x+9cos⁡x)−(9sin⁡x−xcos⁡x)(xcos⁡x−8sin⁡x)(xsin⁡x+9cos⁡x)2\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}
=(8cos⁡x+xsin⁡x)(xsin⁡x+9cos⁡x)−(9sin⁡x−xcos⁡x)(xcos⁡x−8sin⁡x)(xsin⁡x+9cos⁡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}
=8xsin⁡xcos⁡x+72cos⁡2x+x2sin⁡2x+9xsin⁡xcos⁡x−(9xsin⁡xcos⁡x−72sin⁡2x−x2cos⁡2x+8xsin⁡xcos⁡x)(xsin⁡x+9cos⁡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}
=8xsin⁡xcos⁡x+72cos⁡2x+x2sin⁡2x+9xsin⁡xcos⁡x−9xsin⁡xcos⁡x+72sin⁡2x+x2cos⁡2x−8xsin⁡xcos⁡x(xsin⁡x+9cos⁡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}
=x2(sin⁡2x+cos⁡2x)+72(cos⁡2x+sin⁡2x)(xsin⁡x+9cos⁡x)2=x2+72(xsin⁡x+9cos⁡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, ∫x2+72(xsin⁡x+9cos⁡x)2dx=9sin⁡x−xcos⁡xxsin⁡x+9cos⁡x+C\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

9sin⁡x−xcos⁡xxsin⁡x+9cos⁡x+C\frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x} + C

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