Question-168138

Question Number 168138 by Shemson_buo last updated on 04/Apr/22

Answered by MJS_new last updated on 04/Apr/22

∫(dx/( (√(1−2x−x^2 ))))= [t=arcsin ((x+1)/( (√2))) → dx=(√(1−2x−x^2 ))dt] =∫dt=t=arcsin ((x+1)/( (√2))) +C

$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\rightarrow\:{dx}=\sqrt{\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} }{dt}\right] \\ $$$$=\int{dt}={t}=\mathrm{arcsin}\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+{C} \\ $$

Commented by MJS_new last updated on 04/Apr/22

in 3 steps ∫(dx/( (√(1−2x−x^2 ))))= [r=x+1 → dx=dr] =∫(dr/( (√(2−r^2 ))))= [s=(r/( (√2))) → dr=(√2)ds] =∫(ds/( (√(1−s^2 ))))= [t=arcsin s → ds=(√(1−s^2 ))dt] =∫dt=t=arcsin s =arcsin (r/( (√2))) = =arcsin ((x+1)/( (√2))) +C

$$\mathrm{in}\:\mathrm{3}\:\mathrm{steps} \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{r}={x}+\mathrm{1}\:\rightarrow\:{dx}={dr}\right] \\ $$$$=\int\frac{{dr}}{\:\sqrt{\mathrm{2}−{r}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{s}=\frac{{r}}{\:\sqrt{\mathrm{2}}}\:\rightarrow\:{dr}=\sqrt{\mathrm{2}}{ds}\right] \\ $$$$=\int\frac{{ds}}{\:\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:{s}\:\rightarrow\:{ds}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }{dt}\right] \\ $$$$=\int{dt}={t}=\mathrm{arcsin}\:{s}\:=\mathrm{arcsin}\:\frac{{r}}{\:\sqrt{\mathrm{2}}}\:= \\ $$$$=\mathrm{arcsin}\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+{C} \\ $$

Answered by Mathspace last updated on 04/Apr/22

I=∫ (dx/( (√(1−2x−x^2 )))) we have 1−2x−x^2 =−(x^2 +2x−1) =−(x^2 +2x+1−2) =2−(x+1)^2 and changement x+1=(√2)sint give I=∫ (((√2)cost dt)/( (√2)cost))=∫ dt +C =t+C=arcsin(((x+1)/( (√2))))+C

$${I}=\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} }} \\ $$$${we}\:{have}\:\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} =−\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$=−\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{2}\right) \\ $$$$=\mathrm{2}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:{and}\:{changement} \\ $$$${x}+\mathrm{1}=\sqrt{\mathrm{2}}{sint}\:{give} \\ $$$${I}=\int\:\:\frac{\sqrt{\mathrm{2}}{cost}\:{dt}}{\:\sqrt{\mathrm{2}}{cost}}=\int\:{dt}\:+{C} \\ $$$$={t}+{C}={arcsin}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{C} \\ $$

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