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Question Number 132610 by liberty last updated on 15/Feb/21
Ω=∫ ((sin^2 (x))/(1+sin^2 (x))) dx
$$\Omega=\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$
Answered by Dwaipayan Shikari last updated on 15/Feb/21
∫((sin^2 x)/(1+sin^2 x))dx=x−∫(1/(1+sin^2 x))dx =x−∫((cosec^2 x)/(cot^2 x+2))dx=x−(1/( (√2)))tan^(−1) ((cotx)/( (√2)))+C
$$\int\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx}={x}−\int\frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$={x}−\int\frac{{cosec}^{\mathrm{2}} {x}}{{cot}^{\mathrm{2}} {x}+\mathrm{2}}{dx}={x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \frac{{cotx}}{\:\sqrt{\mathrm{2}}}+{C} \\ $$
Answered by mathmax by abdo last updated on 15/Feb/21
Φ =∫ ((sin^2 x)/(1+sin^2 x))dx ⇒Φ =∫ (((1−cos(2x))/2)/(1+((1−cos(2x))/2)))dx =∫ ((1−cos(2x))/(3−cos(2x)))dx =_(2x=t) (1/2) ∫ ((1−cost)/(3−cost))dt =_(tan((t/2))=y) (1/2)∫ ((1−((1−y^2 )/(1+y^2 )))/(3−((1−y^2 )/(1+y^2 ))))((2dy)/(1+y^2 )) =∫ ((1+y^2 −1+y^2 )/(3+3y^2 −1+y^2 ))dy =∫ ((2y^2 )/(2+4y^2 ))dy =∫ (y^2 /(1+2y^2 ))dy =(1/2)∫ ((2y^2 +1−1)/(1+2y^2 ))dy =(y/2)−(1/2)∫ (dy/(1+2y^2 ))(→t=(√2)y) =(y/2)−(1/2)∫ (dt/( (√2)(1+t^2 ))) =(y/2)−(1/(2(√2))) arctan((√2)y)+C =(1/2)tan(x)−(1/(2(√2)))arctan((√2)tan(x))+C
$$\Phi\:=\int\:\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\Rightarrow\Phi\:=\int\:\:\frac{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx} \\ $$$$=\int\:\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{3}−\mathrm{cos}\left(\mathrm{2x}\right)}\mathrm{dx}\:=_{\mathrm{2x}=\mathrm{t}} \frac{\mathrm{1}}{\mathrm{2}}\:\:\:\int\:\:\frac{\mathrm{1}−\mathrm{cost}}{\mathrm{3}−\mathrm{cost}}\mathrm{dt} \\ $$$$=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{y}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}{\mathrm{3}−\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\frac{\mathrm{2dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{1}+\mathrm{y}^{\mathrm{2}} −\mathrm{1}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}+\mathrm{3y}^{\mathrm{2}} −\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\mathrm{dy} \\ $$$$=\int\:\:\frac{\mathrm{2y}^{\mathrm{2}} }{\mathrm{2}+\mathrm{4y}^{\mathrm{2}} }\mathrm{dy}\:=\int\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{2y}^{\mathrm{2}} }\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2y}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+\mathrm{2y}^{\mathrm{2}} }\mathrm{dy} \\ $$$$=\frac{\mathrm{y}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{2y}^{\mathrm{2}} }\left(\rightarrow\mathrm{t}=\sqrt{\mathrm{2}}\mathrm{y}\right) \\ $$$$=\frac{\mathrm{y}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{y}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{y}\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\left(\mathrm{x}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{tan}\left(\mathrm{x}\right)\right)+\mathrm{C} \\ $$

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