Using a suitable substitution, find
$$\int \frac{x \sin x \cos x}{(x-\sin x \cos x)^2}dx.$$
I'm really struggling to find a substitution that will help me even get started.
2026-05-16 05:39:42.1778909982
Difficult Integration by Substitution
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We can rewrite the integrand as $$ \frac{x\sec^2{x}\tan{x}}{(x\sec^2{x}-\tan{x})^2} $$ by dividing numerator and denominator by $\cos^4{x}$. Now try putting $ u=x\sec^2{x}-\tan{x} $, so $du = 2x\sec^2{x}\tan{x}+\sec^2{x} - \sec^2{x} = 2x\sec^2{x}\tan{x} $. Hence the integral becomes $ \frac{1}{2} \int du/u^2$.