Question-144059

Question Number 144059 by 0731619 last updated on 21/Jun/21

Answered by Olaf_Thorendsen last updated on 21/Jun/21

F(x) = ∫((sin^2 x)/(b^2 cos^2 x+a^2 )) dx F(x) = ∫((tan^2 x)/(b^2 +(a^2 /(cos^2 x)))) dx F(x) = ∫((tan^2 x)/(b^2 +a^2 (1+tan^2 x))) dx Let t = tanx F(t) = ∫(t^2 /(b^2 +a^2 (1+t^2 ))).(dt/(1+t^2 )) F(t) = ∫((((a^2 +b^2 )/b^2 )/(a^2 t^2 +a^2 +b^2 ))−((1/b^2 )/(1+t^2 )))dt F(t) = ((a^2 +b^2 )/(a^2 b^2 ))∫(dt/(t^2 +((a^2 +b^2 )/a^2 )))−(1/b^2 )∫(dt/(1+t^2 )) F(t) = ((a^2 +b^2 )/(a^2 b^2 ))((1/( (√((a^2 +b^2 )/a^2 ))))arctan(t/( (√((a^2 +b^2 )/a^2 ))))) −(1/b^2 )arctant+C F(t) = ((√(a^2 +b^2 ))/(ab^2 ))arctan(((at)/( (√(a^2 +b^2 )))))−(1/b^2 )arctant+C F(x) = ((√(a^2 +b^2 ))/(ab^2 ))arctan(((a.tanx)/( (√(a^2 +b^2 )))))−(x/b^2 )+C

$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{sin}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{tan}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} {x}}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{tan}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)}\:{dx} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}{x} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\int\frac{{t}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}.\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left({t}\right)\:=\:\int\left(\frac{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\int\frac{{dt}}{{t}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}\mathrm{arctan}\frac{{t}}{\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}\right) \\ $$$$−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\mathrm{arctan}{t}+\mathrm{C} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{{at}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\mathrm{arctan}{t}+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{{a}.\mathrm{tan}{x}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)−\frac{{x}}{{b}^{\mathrm{2}} }+\mathrm{C} \\ $$

Commented by 0731619 last updated on 21/Jun/21

$${thanks} \\ $$

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