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Question Number 97417 by MJS last updated on 08/Jun/20

$$\int \frac{x}{a+{\sin }^{2}x}dx=?$$

Answered by MJS last updated on 08/Jun/20

$$\mathrm{just}\mathrm{found}\mathrm{a}\mathrm{path}:$$$$\sin x=\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{2\mathrm{i}}$$$$\Rightarrow$$$$\int \frac{x}{a+{\sin }^{2}x}dx=4\int \frac{x{\mathrm{e}}^{2\mathrm{i}x}}{4a{\mathrm{e}}^{2\mathrm{i}x}-{({\mathrm{e}}^{2\mathrm{i}x}-1)}^2}dx=$$$$[t={\mathrm{e}}^{2\mathrm{i}x}-1\to dx=-\frac{\mathrm{i}}{{2\mathrm{e}}^{2\mathrm{i}x}}dx]$$$$=\int \frac{\ln (t+1)}{t^2-4at-4a}dt$$$$\mathrm{now}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{factorize}\mathrm{and}\mathrm{decompose}$$$$\mathrm{then}\mathrm{substitute}\mathrm{to}\mathrm{get}\mathrm{integrals}\mathrm{of}\mathrm{the}\mathrm{shape}$$$$\int \frac{\ln (1-u)}{u}du=-{\mathrm{Li}}_2\left(u\right)$$$$\mathrm{I}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{only}\mathrm{hard}\mathrm{to}\mathrm{type}...$$

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