<s>[INST] Trigonometric integrals How do I evaluate this indefinite integral ?  Integral $$\int\frac{x^2+n(n-1)}{(x\sin(x)+n\cos(x))^2}dx$$ What type of integral is it ? Is there any intuition involved in the approach to solve it?
Edit: The complete term in denominator has 2 as exponent. There was a typo previously.
 [/INST] $\bf{My\; Solution:}$ Using $$\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $$\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$$ and $$\displaystyle \cos \phi = \frac{n}{\sqrt{x^2+n^2}}$$ and $$\displaystyle \tan \phi = \frac{x}{n}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{n}\right)$$
So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx$$
Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=y.$$ Then $$\displaystyle \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dy$$
So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathbb{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)+\mathcal{C}$$
So $$\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx = \left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathcal{C}.$$
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