In the following integration
$$\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x$$
I tried alot. But does not get any proper start.
Can anybody provide me a hint.
Hint in integration $\int\frac{x^{2}}{\left(x\cos x-\sin x \right )\left( x\sin x+\cos x \right )}\,\mathrm{d}x$
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calculus
integration
indefinite-integrals
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0we had this question more the once here... – 2017-01-07
2 Answers
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Hint. One may observe the following trick $$ \begin{align} \frac{x^2}{(x \cos x-\sin x)(x\sin x+\cos x)}&=\frac{x\cos x(x \cos x-\sin x)+x\sin x(x\sin x+\cos x)}{(x \cos x-\sin x)(x\sin x+\cos x)} \\\\&=\frac{x\cos x}{x\sin x+\cos x}+\frac{x \sin x}{x \cos x-\sin x} \\\\&=\frac{(x\sin x+\cos x)'}{(x\sin x+\cos x)}-\frac{(x \cos x-\sin x)'}{(x \cos x-\sin x)} \end{align} $$ then one may conclude.
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Notice $$x^{2}=x^2\left ( \sin^2x+\cos^2x \right )$$ then \begin{align*} \int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x&=\int \frac{x^2\left ( \sin^2x+\cos^2x \right )}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x\\ &=\int \frac{x\cos x}{x\sin x+\cos x}\, \mathrm{d}x+\int \frac{x\sin x}{x\cos x-\sin x}\, \mathrm{d}x\\ &=\ln\left | x\sin x+\cos x \right |-\ln\left | x\cos x-\sin x \right |+C \end{align*}