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xtan-2-x-dx-




Question Number 11741 by uni last updated on 30/Mar/17
∫xtan^2  x dx=?
$$\int\mathrm{xtan}^{\mathrm{2}} \:\mathrm{x}\:\mathrm{dx}=? \\ $$
Answered by mrW1 last updated on 30/Mar/17
∫xtan^2  x dx=∫tan^2  x d((x^2 /2))  =((x^2 tan^2  x)/2)−∫x^2 ((sin x)/(cos^3  x))dx  =((x^2 tan^2  x)/2)+(1/2)∫x^2 d((1/(cos^2  x)))  =((x^2 tan^2  x)/2)+(x^2 /(2cos^2  x))−∫(x/(cos^2  x))dx  =((x^2 (1+sin^2  x))/(2cos^2  x))−∫xd(tan x)  =((x^2 (1+sin^2  x))/(2cos^2  x))−xtan x+∫tan xdx  =((x^2 (1+sin^2  x))/(2cos^2  x))−xtan x−ln ∣cos x∣+C
$$\int{x}\mathrm{tan}^{\mathrm{2}} \:{x}\:{dx}=\int\mathrm{tan}^{\mathrm{2}} \:{x}\:{d}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\frac{{x}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{2}}−\int{x}^{\mathrm{2}} \frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int{x}^{\mathrm{2}} {d}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\right) \\ $$$$=\frac{{x}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2cos}^{\mathrm{2}} \:{x}}−\int\frac{{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{\mathrm{2cos}^{\mathrm{2}} \:{x}}−\int{xd}\left(\mathrm{tan}\:{x}\right) \\ $$$$=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{\mathrm{2cos}^{\mathrm{2}} \:{x}}−{x}\mathrm{tan}\:{x}+\int\mathrm{tan}\:{xdx} \\ $$$$=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{\mathrm{2cos}^{\mathrm{2}} \:{x}}−{x}\mathrm{tan}\:{x}−\mathrm{ln}\:\mid\mathrm{cos}\:{x}\mid+{C} \\ $$

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