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csc-2019-x-sec-5-x-tan-2-x-dx-




Question Number 60938 by aliesam last updated on 27/May/19
∫((csc^(2019) (x))/(sec^5 (x))) tan^2 (x) dx
$$\int\frac{{csc}^{\mathrm{2019}} \left({x}\right)}{{sec}^{\mathrm{5}} \left({x}\right)}\:{tan}^{\mathrm{2}} \left({x}\right)\:{dx} \\ $$
Commented by Prithwish sen last updated on 27/May/19
∫((cos^3 x)/(sin^(2017) x))dx =∫(((1−sin^2 x)cosx)/(sin^(2017) x)) dx  put sinx = v  then cosx dx =dv  =∫(((1−v^2 )dv)/v^(2017) )  =∫(v^(−2017) −v^(−2015) ) dv
$$\int\frac{\mathrm{cos}^{\mathrm{3}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2017}} \mathrm{x}}\mathrm{dx}\:=\int\frac{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2017}} \mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{put}\:\mathrm{sinx}\:=\:\mathrm{v} \\ $$$$\mathrm{then}\:\mathrm{cosx}\:\mathrm{dx}\:=\mathrm{dv} \\ $$$$=\int\frac{\left(\mathrm{1}−\mathrm{v}^{\mathrm{2}} \right)\mathrm{dv}}{\mathrm{v}^{\mathrm{2017}} } \\ $$$$=\int\left(\mathrm{v}^{−\mathrm{2017}} −\mathrm{v}^{−\mathrm{2015}} \right)\:\mathrm{dv} \\ $$$$ \\ $$

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