1-1-9x-2-dx-

Question Number 145777 by Engr_Jidda last updated on 08/Jul/21

$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{dx} \\ $$

Answered by ArielVyny last updated on 08/Jul/21

∫((√(1−9x^2 ))/(1−9x^2 ))dx=∫((√(1−9x^2 ))/((1−3x)(1+3x)))dx (a/(1−3x))+(b/(1+3x))=(1/(1−9x^2 )) ((a+3ax+b−3bx)/(1−9x^2 ))=(1/(1−9x^2 )) ((a+b+x(3a−3b))/(1−9x^2 ))=(1/(1−9x^2 )) { ((a+b=1)),((a=b)) :}→a=b=(1/2) (1/2)∫(((√(1−9x^2 ))/(1−3x))+((√(1−9x^2 ))/(1+3x)))dx posons 3x=sint→3dx=costdt (1/6)∫((cos^2 t)/(1−sint))dt+((cos^2 t)/(1+sint))dt (1/6)∫(1+sint)+(1−sint)=(1/6)∫2dt=(1/3)t ∫(1/( (√(1−9x^2 ))))=(1/3)arcsin(3x)+cte

$$\int\frac{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }{dx}=\int\frac{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{\left(\mathrm{1}−\mathrm{3}{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)}{dx} \\ $$$$\frac{{a}}{\mathrm{1}−\mathrm{3}{x}}+\frac{{b}}{\mathrm{1}+\mathrm{3}{x}}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$\frac{{a}+\mathrm{3}{ax}+{b}−\mathrm{3}{bx}}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$\frac{{a}+{b}+{x}\left(\mathrm{3}{a}−\mathrm{3}{b}\right)}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$\begin{cases}{{a}+{b}=\mathrm{1}}\\{{a}={b}}\end{cases}\rightarrow{a}={b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{3}{x}}+\frac{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{3}{x}}\right){dx} \\ $$$${posons}\:\mathrm{3}{x}={sint}\rightarrow\mathrm{3}{dx}={costdt} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{cos}^{\mathrm{2}} {t}}{\mathrm{1}−{sint}}{dt}+\frac{{cos}^{\mathrm{2}} {t}}{\mathrm{1}+{sint}}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\left(\mathrm{1}+{sint}\right)+\left(\mathrm{1}−{sint}\right)=\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{2}{dt}=\frac{\mathrm{1}}{\mathrm{3}}{t} \\ $$$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{3}}{arcsin}\left(\mathrm{3}{x}\right)+{cte} \\ $$

Answered by mathmax by abdo last updated on 08/Jul/21

I=∫ (dx/( (√(1−9x^2 )))) ⇒I=_(3x=sint) ∫ ((cost)/(3cost))dt =(1/3)t +K =(1/3)arcsin(3x)+K

$$\mathrm{I}=\int\:\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{9x}^{\mathrm{2}} }}\:\Rightarrow\mathrm{I}=_{\mathrm{3x}=\mathrm{sint}} \:\:\int\:\:\frac{\mathrm{cost}}{\mathrm{3cost}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}\:+\mathrm{K} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\left(\mathrm{3x}\right)+\mathrm{K} \\ $$

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