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Question Number 9403 by alfat123 last updated on 05/Dec/16

Answered by ridwan balatif last updated on 05/Dec/16

$${\int }_0^{1}5\mathrm{x}{(1-\mathrm{x})}^6\mathrm{dx}=\mathrm{u}.\mathrm{v}-\int \mathrm{vdu}$$$$\mathrm{misal}:\mathrm{u}=5\mathrm{x}\to \mathrm{du}=5\mathrm{dx}$$$$\mathrm{dv}={(1-\mathrm{x})}^6\mathrm{dx}$$$$\int \mathrm{dv}=\int {(1-\mathrm{x})}^6\mathrm{dx}$$$$\mathrm{v}=-\frac{1}{7}{(1-\mathrm{x})}^7$$$$\mathrm{maka},{\int }_0^{1}5\mathrm{x}{(1-\mathrm{x})}^6={(\left(5\mathrm{x}\right).\left(\frac{-1}{7}{(1-\mathrm{x})}^7\right)-(\int -\frac{1}{7}{(1-\mathrm{x})}^{7}5\mathrm{dx}))}_0^1$$$$={(-\frac{5\mathrm{x}}{7}{(1-\mathrm{x})}^7-\frac{5}{56}{(1-\mathrm{x})}^8)}_0^1$$$$=-{\left(\frac{5}{7}{(1-\mathrm{x})}^7(\mathrm{x}+\frac{1}{8}(1-\mathrm{x}))\right)}_0^1$$$$=-{\left(\frac{5}{56}{(1-\mathrm{x})}^7(7\mathrm{x}+1)\right)}_0^1$$$$=\frac{5}{56}$$$$$$$$$$

Answered by ridwan balatif last updated on 05/Dec/16

$$\int {\cos }^3\mathrm{xdx}=\int \left(\mathrm{cosx}\right){\cos }^2\mathrm{xdx}$$$$=\int \mathrm{cosx}(1-{\sin }^2\mathrm{x})\mathrm{dx}$$$$=\int \mathrm{cosdx}-\int {\sin }^2\mathrm{xcosxdx}...(\ast )$$$$\int {\sin }^2\mathrm{xcosxdx}=\mathrm{u}.\mathrm{v}-\int \mathrm{vdu}$$$$\mathrm{misal}:\mathrm{u}={\sin }^2\mathrm{x}\to \mathrm{du}=2\mathrm{sinxcosxdx}$$$$\mathrm{dv}=\mathrm{cosxdx}\to \mathrm{v}=\mathrm{sinx}$$$$\int {\sin }^2\mathrm{xcosxdx}={\sin }^3\mathrm{x}-\int \mathrm{sinx}\left(2\mathrm{sinxcosx}\right)\mathrm{dx}$$$$\int {\sin }^2\mathrm{xcosxdx}={\sin }^3\mathrm{x}-2\int {\sin }^2\mathrm{xcosxdx}$$$$3\int {\sin }^2\mathrm{xcosxdx}={\sin }^3\mathrm{x}$$$$\int {\sin }^2\mathrm{xcosxdx}=\frac{1}{3}{\sin }^3\mathrm{x}+\mathrm{C}...(\ast \ast )$$$$\mathrm{substitusi}\mathrm{nilai}\mathrm{dari}\mathrm{persamaan}(\ast \ast )\mathrm{ke}\mathrm{persamaan}(\ast )$$$$\int {\cos }^3\mathrm{xdx}=\int \mathrm{cosxdx}-\int {\sin }^2\mathrm{xcosxdx}$$$$=\mathrm{sinx}-(\frac{1}{3}{\sin }^3\mathrm{x}+\mathrm{C})$$$$=\mathrm{sinx}-\frac{1}{3}{\sin }^3\mathrm{x}+\mathrm{C}$$

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