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x-2-1-x-2-dx-




Question Number 9467 by tawakalitu last updated on 09/Dec/16
∫x^2 ((√(1 − x^2 ))) dx
$$\int\mathrm{x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} }\right)\:\mathrm{dx} \\ $$
Answered by mrW last updated on 09/Dec/16
x=sin t  dx=cos t dt  I=∫x^2 ((√(1 − x^2 ))) dx=∫sin^2 t×cos^2 t dt  =∫(1/4)(2sin t×cos t)^2  dt  =(1/8)∫sin^2 (2t) d(2t)  =(1/8)[((2t)/2)−(1/4)sin (2t)]+C  =(1/(32))[4t−sin (2t)]+C  t=sin^(−1) x  sin (2t)=2sin tcos t=2x(√(1−x^2 ))  I=(1/(16))[2sin^(−1) x−x(√(1−x^2 ))]+C
$$\mathrm{x}=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{dx}=\mathrm{cos}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{I}=\int\mathrm{x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} }\right)\:\mathrm{dx}=\int\mathrm{sin}\:^{\mathrm{2}} \mathrm{t}×\mathrm{cos}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2sin}\:\mathrm{t}×\mathrm{cos}\:\mathrm{t}\right)^{\mathrm{2}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2t}\right)\:\mathrm{d}\left(\mathrm{2t}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{2t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\left(\mathrm{2t}\right)\right]+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left[\mathrm{4t}−\mathrm{sin}\:\left(\mathrm{2t}\right)\right]+\mathrm{C} \\ $$$$\mathrm{t}=\mathrm{sin}\:^{−\mathrm{1}} \mathrm{x} \\ $$$$\mathrm{sin}\:\left(\mathrm{2t}\right)=\mathrm{2sin}\:\mathrm{tcos}\:\mathrm{t}=\mathrm{2x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{16}}\left[\mathrm{2sin}\:^{−\mathrm{1}} \mathrm{x}−\mathrm{x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right]+\mathrm{C} \\ $$
Commented by tawakalitu last updated on 09/Dec/16
Wow. God bless you sir.
$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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