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Question Number 17167 by Arnab Maiti last updated on 01/Jul/17

∫x^2 sin^(−1) 3x dx

$$\int\mathrm{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \mathrm{3x}\:\mathrm{dx} \\ $$

Answered by sma3l2996 last updated on 01/Jul/17

I=∫x^2 sin^(−1) (3x)dx I=(1/3)x^3 sin^(−1) (3x)−∫(x^3 /(√(1−9x^2 )))dx+c let : t=(√(1−9x^2 ))⇒dt=((−9xdx)/(√(1−9x^2 ))) x^2 dt=((−9x^3 dx)/(√(1−9x^2 )))⇔(1/9)(1−t^2 )dt=((−9x^3 )/(√(1−9x^2 )))dx (x^3 /(√(1−9x^2 )))dx=−(1/(81))(1−t^2 )dt I=(1/3)x^3 sin^(−1) (3x)+(1/(81))∫(1−t^2 )dt+c I=(1/3)x^3 sin^(−1) (3x)+(1/(81))(t−(1/3)t^3 )+C I=(1/3)x^3 sin^(−1) (3x)+(1/(243))(√(1−9x^2 ))(2+9x^2 )+C

$${I}=\int{x}^{\mathrm{2}} {sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)−\int\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{dx}+{c} \\ $$$${let}\::\:{t}=\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }\Rightarrow{dt}=\frac{−\mathrm{9}{xdx}}{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }} \\ $$$${x}^{\mathrm{2}} {dt}=\frac{−\mathrm{9}{x}^{\mathrm{3}} {dx}}{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}=\frac{−\mathrm{9}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }}{dx}=−\frac{\mathrm{1}}{\mathrm{81}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)+\frac{\mathrm{1}}{\mathrm{81}}\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}+{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)+\frac{\mathrm{1}}{\mathrm{81}}\left({t}−\frac{\mathrm{1}}{\mathrm{3}}{t}^{\mathrm{3}} \right)+{C} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)+\frac{\mathrm{1}}{\mathrm{243}}\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }\left(\mathrm{2}+\mathrm{9}{x}^{\mathrm{2}} \right)+{C} \\ $$

Commented by Arnab Maiti last updated on 02/Jul/17

Thank you very much.

$$\mathcal{T}\mathfrak{hank}\:\mathfrak{you}\:\mathfrak{very}\:\mathfrak{much}. \\ $$

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