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Question Number 10188 by ridwan balatif last updated on 29/Jan/17

Commented by ridwan balatif last updated on 29/Jan/17

∫(√(1−tan^2 2x+tan^4 2x−tan^6 2x+...))dx=...? 1−tan^2 2x+tan^4 2x−tan^6 2x+...= U_1 =1 r=−tan^2 2x S_∞ =(U_1 /(1−r))=(1/(1+tan^2 2x))=(1/(sec^2 2x))=cos^2 2x S_∞ =cos^2 2x (√(1−tan^2 2x+tan^4 2x−tan^6 2x+...))=(√(cos^2 2x))=cos2x so, ∫cos2x dx=(1/2)sin2x + C is it true my way?

$$\int\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{2x}+\mathrm{tan}^{\mathrm{4}} \mathrm{2x}−\mathrm{tan}^{\mathrm{6}} \mathrm{2x}+...}\mathrm{dx}=...? \\ $$$$\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{2x}+\mathrm{tan}^{\mathrm{4}} \mathrm{2x}−\mathrm{tan}^{\mathrm{6}} \mathrm{2x}+...=\: \\ $$$$\mathrm{U}_{\mathrm{1}} =\mathrm{1}\:\:\mathrm{r}=−\mathrm{tan}^{\mathrm{2}} \mathrm{2x} \\ $$$$\mathrm{S}_{\infty} =\frac{\mathrm{U}_{\mathrm{1}} }{\mathrm{1}−\mathrm{r}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{2x}}=\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \mathrm{2x}}=\mathrm{cos}^{\mathrm{2}} \mathrm{2x} \\ $$$$\mathrm{S}_{\infty} =\mathrm{cos}^{\mathrm{2}} \mathrm{2x} \\ $$$$\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{2x}+\mathrm{tan}^{\mathrm{4}} \mathrm{2x}−\mathrm{tan}^{\mathrm{6}} \mathrm{2x}+...}=\sqrt{\mathrm{cos}^{\mathrm{2}} \mathrm{2x}}=\mathrm{cos2x} \\ $$$$\mathrm{so},\:\int\mathrm{cos2x}\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2x}\:+\:\mathrm{C} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{true}\:\mathrm{my}\:\mathrm{way}? \\ $$

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