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Question Number 119970 by bramlexs22 last updated on 28/Oct/20

∫ (dx/(x^2 (√(25−x^2 )))) ?

$$\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:? \\ $$

Answered by bemath last updated on 28/Oct/20

∫ (dx/(x^3 (√(((25)/x^2 )−1)))) dx = ∫ (x^(−3) /( (√(25x^(−2) −1)))) dx letting m = 25x^(−2) −1⇒dm = −50x^(−3) dx ∫ ((−(dm/(50)))/( (√m))) = −(1/(50))∫ m^(−(1/2)) dm = −(1/(25))(√m) + c = −(1/(25))(√(25x^(−2) −1))+c =−((√(25−x^2 ))/(25x)) + c

$$\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\frac{\mathrm{25}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{25}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$${letting}\:{m}\:=\:\mathrm{25}{x}^{−\mathrm{2}} −\mathrm{1}\Rightarrow{dm}\:=\:−\mathrm{50}{x}^{−\mathrm{3}} {dx} \\ $$$$\int\:\frac{−\frac{{dm}}{\mathrm{50}}}{\:\sqrt{{m}}}\:=\:−\frac{\mathrm{1}}{\mathrm{50}}\int\:{m}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dm} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{25}}\sqrt{{m}}\:+\:{c}\:=\:−\frac{\mathrm{1}}{\mathrm{25}}\sqrt{\mathrm{25}{x}^{−\mathrm{2}} −\mathrm{1}}+{c} \\ $$$$=−\frac{\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}{\mathrm{25}{x}}\:+\:{c}\: \\ $$

Answered by Dwaipayan Shikari last updated on 28/Oct/20

∫(dx/(25sin^2 θ(√(25−25sin^2 θ)))) x=5sinθ⇒1=5cosθ(dθ/dx) =∫((5cosθdθ)/(25sin^2 θ.5cosθ)) =(1/(25))∫(1/(sin^2 θ)) =−(1/(25))(tanθ)^(−1) =−(1/(25))(.((sinθ)/( (√(1−sin^2 θ)))))^(−1) =−(1/(25))(.((5x)/(5(√(25−x^2 )))))^(−1) =−(1/(25)).((x/( (√(25−x^2 )))))^(−1) +C =−(1/(25))(((√(25−x^2 ))/x))+C

$$\int\frac{{dx}}{\mathrm{25}{sin}^{\mathrm{2}} \theta\sqrt{\mathrm{25}−\mathrm{25}{sin}^{\mathrm{2}} \theta}}\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{5}{sin}\theta\Rightarrow\mathrm{1}=\mathrm{5}{cos}\theta\frac{{d}\theta}{{dx}} \\ $$$$=\int\frac{\mathrm{5}{cos}\theta{d}\theta}{\mathrm{25}{sin}^{\mathrm{2}} \theta.\mathrm{5}{cos}\theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\int\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{25}}\left({tan}\theta\right)^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{25}}\left(.\frac{{sin}\theta}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}}\right)^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{25}}\left(.\frac{\mathrm{5}{x}}{\mathrm{5}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\right)^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{25}}.\left(\frac{{x}}{\:\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\right)^{−\mathrm{1}} +{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}{{x}}\right)+{C} \\ $$

Commented by bemath last updated on 28/Oct/20

cot θ = ((cos θ)/( (√(1−cos^2 θ))))

$$\mathrm{cot}\:\theta\:=\:\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta}} \\ $$

Commented by Dwaipayan Shikari last updated on 28/Oct/20

I suppose it as tanθ typo Thanking for correction

$${I}\:{suppose}\:{it}\:{as}\:{tan}\theta\:\:\:{typo} \\ $$$${Thanking}\:{for}\:{correction} \\ $$

Answered by Bird last updated on 28/Oct/20

I =∫ x^(−2) (25−x^2 )^(−(1/2)) dx by parts I =−(1/x)(25−x^2 )^(−(1/2)) −∫−(1/x)(−(1/2))(−2x)(25−x^2 )^(−(3/2)) =−(1/(x(√(25−x^2 ))))+∫ (dx/((25−x^2 )(√(25−x^2 )))) ∫ (dx/((25−x^2 )(√(25−x^2 ))))=_(x=5sint) =∫ ((5cost dt)/(25 cos^2 t .5.cost)) =(1/(25))∫ (dt/((1+cos(2t))/2)) =(2/(25))∫ (dt/(1+cos(2t))) =_(tant =u) (2/(25))∫ (du/((1+u^2 )(1+((1−u^2 )/(1+u^2 ))))) =(2/(25))∫ (du/(1+u^2 +1−u^2 )) =(1/(25))∫ du =(u/(25))+c =tan( arcsin((x/5)))+c ⇒ I =−(1/(x(√(25−x^2 )))) +tan(arcsin((x/5))+C

$${I}\:=\int\:{x}^{−\mathrm{2}} \left(\mathrm{25}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx}\:{by}\:{parts} \\ $$$${I}\:=−\frac{\mathrm{1}}{{x}}\left(\mathrm{25}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} −\int−\frac{\mathrm{1}}{{x}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\mathrm{2}{x}\right)\left(\mathrm{25}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{{x}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}+\int\:\:\frac{{dx}}{\left(\mathrm{25}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }} \\ $$$$\int\:\:\:\frac{{dx}}{\left(\mathrm{25}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}=_{{x}=\mathrm{5}{sint}} \\ $$$$=\int\:\:\:\frac{\mathrm{5}{cost}\:{dt}}{\mathrm{25}\:{cos}^{\mathrm{2}} {t}\:.\mathrm{5}.{cost}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\int\:\:\frac{{dt}}{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\frac{\mathrm{2}}{\mathrm{25}}\int\:\:\frac{{dt}}{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)} \\ $$$$=_{{tant}\:={u}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{25}}\int\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}\int\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} +\mathrm{1}−{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{25}}\int\:{du} \\ $$$$=\frac{{u}}{\mathrm{25}}+{c}\:={tan}\left(\:{arcsin}\left(\frac{{x}}{\mathrm{5}}\right)\right)+{c}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{{x}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:+{tan}\left({arcsin}\left(\frac{{x}}{\mathrm{5}}\right)+{C}\right. \\ $$

Commented by Bird last updated on 28/Oct/20

I =−(1/(x(√(25−x^2 )))) +(1/(25))arctan(arcsin((x/5))) +C

$${I}\:=−\frac{\mathrm{1}}{{x}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\mathrm{25}}{arctan}\left({arcsin}\left(\frac{{x}}{\mathrm{5}}\right)\right)\:+{C} \\ $$

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