Integral Problem 1:

thr19November 28, 2019November 28, 2019CalculusCalculus, integral, integrals, substitution

$\LARGE{\int \sqrt{\dfrac{x}{1-x^3}} dx =?}$

Solution

Let $x^{3/2} = u$ , therefore $\dfrac{3}{2} \sqrt{x} dx = du \Rightarrow \sqrt{x}dx=\dfrac{2}{3} du$ .

So, we can re-write integral as,

$\int \dfrac{2}{3} \dfrac{du}{\sqrt{1-u^2}} = \dfrac{2}{3} \int \dfrac{du}{\sqrt{1-u^2}} = \dfrac{2}{3} \arcsin (u) + const.$

And by substituting $u$ with $x^{3/2}$ we get

$\dfrac{2}{3} \arcsin (x^{3/2}) + const.$

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