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Question Number 27596 by abdo imad last updated on 10/Jan/18

$$find\int {}^3\sqrt{x^2-x^3}dx$$

Commented by abdo imad last updated on 28/Jan/18

$$I=\int {}^3\sqrt{x^3(\frac{1}{x}-1)}dx=\int x\sqrt{\frac{1}{x}-1}dxletusethech.$$$$\sqrt{\frac{1}{x}-1}=t\Rightarrow \frac{1}{x}-1=t^2\Rightarrow \frac{1}{x}=1+t^2\Rightarrow x=\frac{1}{1+t^2}$$$$I=\int \frac{1}{1+t^2}t.\frac{-2t}{{(1+t^2)}^2}dt=-2\int \frac{t^2}{{(1+t^2)}^3}dt$$$$=-2\int \frac{1+t^2-1}{{(1+t^2)}^3}dt=-2\int \frac{dt}{{(1+t^2)}^2}+2\int \frac{dt}{{(1+t^2)}^3}but$$$$thech.t=tan\theta give$$$$I=-2\int \frac{1+{tan}^2\theta }{{(1+{tan}^2\theta )}^2}d\theta +2\int \frac{1+{tan}^2\theta }{{(1+{tan}^2\theta )}^3}d\theta$$$$=-2\int \frac{d\theta }{1+{tan}^2\theta }+2\int \frac{d\theta }{{(1+{tan}^2\theta )}^2}$$$$=-2\int {cos}^2\theta d\theta +2\int {cos}^4\theta d\theta butwehave$$$$\int {cos}^2\theta d\theta =\int \frac{1+cos\left(2\theta \right)}{2}d\theta =\frac{1}{2}\theta +\frac{1}{4}sin\left(2\theta \right)$$$$\int {cos}^4\theta d\theta =\int {\left(\frac{1+cos\left(2\theta \right)}{2}\right)}^{2}d\theta =\frac{1}{4}\int (1+2cos\left(2\theta \right)+\frac{1+cos\left(4\theta \right)}{2})d\theta$$$$=\frac{1}{4}\theta +\frac{1}{4}sin\left(2\theta \right)+\frac{1}{8}\theta +\frac{1}{32}sin\left(4\theta \right)$$$$=\frac{3}{8}\theta +\frac{1}{4}sin\left(2\theta \right)+\frac{1}{32}sin\left(4\theta \right)so$$$$I=-\theta -\frac{1}{2}sin\left(2\theta \right)+\frac{3}{4}\theta +\frac{1}{2}sin\left(2\theta \right)+\frac{1}{16}sin\left(4\theta \right)$$$$I=\frac{-1}{4}\theta +\frac{1}{16}sin\left(4\theta \right)with\theta =arctan\left(\sqrt{\frac{1}{x}-1}\right).$$$$$$

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