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Question Number 128385 by n0y0n last updated on 06/Jan/21

$${\int }_1^{\pi }\left|\begin{array}{ccc}{\mathrm{x}}^3& \mathrm{lnx}& \mathrm{sinx}\\ {3\mathrm{x}}^2& \frac{1}{\mathrm{x}}& \mathrm{cosx}\\ 6& {2\mathrm{x}}^{-3}& -\mathrm{cosx}\end{array}\right|\mathrm{dx}=?$$

Answered by mathmax by abdo last updated on 06/Jan/21

$$\mathrm{detm}\left(\mathrm{x}\right)={\mathrm{x}}^3\left|\begin{array}{c}\frac{1}{\mathrm{x}}\mathrm{cosx}\\ {2\mathrm{x}}^{-3}-\mathrm{cosx}\end{array}\right|-{3\mathrm{x}}^2\left|\begin{array}{c}\mathrm{lnx}\mathrm{sinx}\\ {2\mathrm{x}}^{-3}-\mathrm{cosx}\end{array}\right|$$$$+6\left|\begin{array}{c}\mathrm{lnx}\mathrm{sinx}\\ \frac{1}{\mathrm{x}}\mathrm{cosx}\end{array}\right|={\mathrm{x}}^3(-\frac{\mathrm{cosx}}{\mathrm{x}}-2\mathrm{cosx}{\mathrm{x}}^{-3})$$$$-{3\mathrm{x}}^2(-\mathrm{lnx}\mathrm{cosx}-2\mathrm{sinx}{\mathrm{x}}^{-3})+6(\mathrm{lnx}\mathrm{cosx}-\frac{\mathrm{sinx}}{\mathrm{x}})$$$$=-{\mathrm{x}}^2\mathrm{cosx}-2\mathrm{cosx}+{3\mathrm{x}}^2\mathrm{lnx}\mathrm{cosx}+\frac{6\mathrm{sinx}}{\mathrm{x}}+6\mathrm{lnx}\mathrm{cosx}-\frac{6\mathrm{sinx}}{\mathrm{x}}$$$$=(-{\mathrm{x}}^2+{3\mathrm{x}}^2\mathrm{lnx}+6\mathrm{lnx})\mathrm{cosx}\Rightarrow$$$${\int }_1^{\pi }\mathrm{detm}\left(\mathrm{x}\right)\mathrm{dx}={\int }_1^{\pi }(-{\mathrm{x}}^2+({3\mathrm{x}}^2+6)\mathrm{lnx})\mathrm{cosx}\mathrm{dx}$$$$=-{\int }_1^{\pi }{\mathrm{x}}^2\mathrm{cosx}\mathrm{dx}+3{\int }_1^{\pi }({\mathrm{x}}^2+2)\mathrm{lnx}\mathrm{cosx}\mathrm{dx}$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{by}\mathrm{parts}...\mathrm{be}\mathrm{continued}...$$

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