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Question Number 125313 by john_santu last updated on 10/Dec/20

$$\beta \left(x\right)=\int \frac{x^3}{\sqrt{1-x^2}}dx$$

Answered by john_santu last updated on 10/Dec/20

$$let\sqrt{1-x^2}=w\Rightarrow x^2=1-w^2$$$$xdx=-wdw$$$$\beta \left(x\right)=\int \frac{(1-w^2)(-wdw)}{w}$$$$=\int (w^2-1)dw=\frac{1}{3}{w}^3-w+c$$$$=\frac{1}{3}w(w^2-3)+c$$$$=\frac{\sqrt{1-x^2}}{3}(1-x^2-3)+c$$$$=\frac{\sqrt{1-x^2}}{3}(-2-x^2)+c$$

Answered by liberty last updated on 10/Dec/20

$$wesubstitute{x}^2=rthenx=r^{\frac{1}{2}}$$$$\beta \left(x\right)=\int \frac{x^3}{\sqrt{1-x^2}}dx=\int r^{\frac{3}{2}}{(1-r)}^{-\frac{1}{2}}\left(\frac{1}{2}{r}^{-\frac{1}{2}}\right)dr$$$$=\frac{1}{2}\int r{(1-r)}^{-\frac{1}{2}}dr$$$$let{(1-r)}^{\frac{1}{2}}=h,r=1-h^2;dr=-2hdh$$$$\beta \left(x\right)=\frac{1}{2}\int (1-h^2)h^{-1}(-2h)dh$$$$=\frac{1}{3}{h}^3-h+c=\frac{1}{3}h(h^2-3)+c$$$$=\frac{\sqrt{1-r}}{3}(-r-2)+c=\frac{\sqrt{1-x^2}}{3}(-x^2-2)+c$$

Answered by Ñï= last updated on 10/Dec/20

Commented by Ñï= last updated on 10/Dec/20

$$x=itan\theta$$

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