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

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

Answered by sma3l2996 last updated on 01/Jul/17

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

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

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

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