Mathematical Reflections 2016, Issue 6 - Problem U395

Problem:
Evaluate $\displaystyle \int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx$.

Proposed by Abdelouahed Hamdi, Doha, Qatar

Solution:Observe that $$\begin{array}{lll} \dfrac{x^2+6}{(x \cos x-3\sin x)^2}&=&\dfrac{x^2(\cos^2 x +\sin^2 x)+6(\cos^2 x + \sin^2 x)+5 x \sin x \cos x - 5x \sin x \cos x}{(x \cos x-3\sin x)^2} \\ &=& \dfrac{(x \cos x-2\sin x)(x \cos x-3 \sin x)+(x \sin x+3 \cos x)(x \sin x+2\cos x)}{(x\cos x-3\sin x)^2}\\&=& \dfrac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2} \\ &=&\left(\dfrac{f(x)}{g(x)}\right)', \end{array}$$ where $f(x)=x\sin x+3\cos x$ and $g(x)=x \cos x -3\sin x$. So, $$\int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx=\dfrac{x\sin x+3\cos x}{x \cos x-3\sin x}+C.$$