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Question Number 128194 by Algoritm last updated on 05/Jan/21

Commented by MJS_new last updated on 05/Jan/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{by}\mathrm{using}$$$$\cos x=\frac{{\mathrm{e}}^{\mathrm{i}x}+{\mathrm{e}}^{-\mathrm{i}x}}{2}=\frac{{\mathrm{e}}^{2\mathrm{i}x}+1}{{2\mathrm{e}}^{\mathrm{i}x}}\Rightarrow$$$$8\int \frac{x{\mathrm{e}}^{3\mathrm{i}x}}{{({\mathrm{e}}^{2\mathrm{i}x}+1)}^3}dx$$$$\mathrm{now}\mathrm{use}t={\mathrm{e}}^{\mathrm{i}x}\mathrm{to}\mathrm{get}$$$$-8\int \frac{t^2\ln t}{{(t^2+1)}^3}dt$$$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{lengthy}\mathrm{and}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{the}\mathrm{time}\mathrm{right}$$$$\mathrm{now}$$

Commented by MJS_new last updated on 06/Jan/21

$$\mathrm{I}\mathrm{get}$$$$-\frac{t}{t^2+1}-(\frac{t(t^2-1)}{{(t^2+1)}^2}+\arctan t)\ln t+\frac{\mathrm{i}}{2}({\mathrm{Li}}_2(-\mathrm{i}t)-{\mathrm{Li}}_2\left(\mathrm{i}t\right))$$$$\mathrm{and}\mathrm{now}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{put}t={\mathrm{e}}^{\mathrm{i}x}.\mathrm{not}\mathrm{sure}\mathrm{how}$$$$\mathrm{to}\mathrm{expand}\mathrm{all}\mathrm{these}\mathrm{functions}...$$

Answered by chengulapetrom last updated on 05/Jan/21

$$\int \frac{x}{{cos}^{3}x}dx=\int {xsec}^{3}xdx$$$$bypartsu=xanddv={sec}^{3}xdx$$$$\frac{du}{dx}=1andv=\int {sec}^{3}xdx$$$$gameover$$

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