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1-cos-x-6-




Question Number 120544 by rubygarfield last updated on 01/Nov/20
∫(1/((cos x)^6 ))=?
$$\int\frac{\mathrm{1}}{\left(\mathrm{cos}\:{x}\right)^{\mathrm{6}} }=? \\ $$
Answered by Dwaipayan Shikari last updated on 01/Nov/20
∫sec^6 xdx  =∫sec^2 x (1+tan^2 x)^2 dx  =∫(1+t^2 )^2 dt       (t=tanx)  =∫1+2t^2 +t^4  dt  =t+(2/3)t^3 +(1/5)t^5 +C  =tanx+(2/3)tan^3 x+(1/5)tan^5 x +C
$$\int{sec}^{\mathrm{6}} {xdx} \\ $$$$=\int{sec}^{\mathrm{2}} {x}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} {dx} \\ $$$$=\int\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} {dt}\:\:\:\:\:\:\:\left({t}={tanx}\right) \\ $$$$=\int\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \:{dt} \\ $$$$={t}+\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{5}}{t}^{\mathrm{5}} +{C} \\ $$$$={tanx}+\frac{\mathrm{2}}{\mathrm{3}}{tan}^{\mathrm{3}} {x}+\frac{\mathrm{1}}{\mathrm{5}}{tan}^{\mathrm{5}} {x}\:+{C} \\ $$
Commented by peter frank last updated on 01/Nov/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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