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Question Number 206200 by Shrodinger last updated on 09/Apr/24

$$\int \frac{xsinx}{1-cosx}dx$$

Answered by Frix last updated on 09/Apr/24

$$\int \frac{x\sin x}{1-\cos x}dx=\mathrm{i}\int \frac{x({\mathrm{e}}^{\mathrm{i}x}+1)}{{\mathrm{e}}^{\mathrm{i}x}-1}dx=$$$$=\mathrm{i}\int xdx+2\mathrm{i}\int \frac{x}{{\mathrm{e}}^{\mathrm{i}x}-1}dx$$$$\mathrm{i}\int xdx=\mathrm{i}\frac{x^2}{2}$$$$2\mathrm{i}\int \frac{x}{{\mathrm{e}}^{\mathrm{i}x}-1}dx\stackrel{t={\mathrm{e}}^{\mathrm{i}x}-1}{=}-2\mathrm{i}\int \frac{\ln (t+1)}{t(t+1)}dt=$$$$=2\mathrm{i}\int \frac{\ln (t+1)}{t+1}dt-2\mathrm{i}\int \frac{\ln (t+1)}{t}dt=$$$$=\mathrm{i}{\ln }^2(t+1)+2\mathrm{i}{\mathrm{Li}}_2\left(t\right)=$$$$=-\mathrm{i}{x}^2+2\mathrm{i}{\mathrm{Li}}_2({\mathrm{e}}^{\mathrm{i}x}-1)$$$$\Rightarrow$$$$\int \frac{x\sin x}{1-\cos x}dx=({2\mathrm{Li}}_2({\mathrm{e}}^{\mathrm{i}x}-1)-\frac{x^2}{2})\mathrm{i}+C$$

Commented by TonyCWX08 last updated on 10/Apr/24

$$LookslikeyoualsoknowtheDilogarithmFunction!$$

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