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Question Number 40745 by Raj Singh last updated on 27/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$x^3=a^3{sin}^2\alpha$$$$3x^{2}dx=a^3.2sin\alpha cos\alpha d\alpha$$$$dx=\frac{2a^{3}sin\alpha cos\alpha }{3{\left(a^3{sin}^2\alpha \right)}^{\frac{2}{3}}}=\frac{2}{3}a.{sin}^{\frac{-1}{3}}\alpha cos\alpha d\alpha$$$$\int \frac{{\left(a^3{sin}^2\alpha \right)}^{\frac{1}{6}}\times \frac{2}{3}{asin}^{\frac{-1}{3}}\alpha cos\alpha d\alpha }{{(a^3-a^3{sin}^2\alpha )}^{\frac{1}{2}}}$$$$\frac{2}{3}\int \frac{a^{\frac{3}{6}+1-\frac{3}{2}}}{}\times {sin}^{\frac{2}{6}-\frac{1}{3}}\alpha d\alpha$$$$\frac{2}{3}\alpha +c$$$$sin\alpha ={\left(\frac{x}{a}\right)}^{\frac{3}{2}}\alpha ={sin}^{-1}\left\{{\left(\frac{x}{a}\right)}^{\frac{3}{2}}\right\}$$$$ansis\frac{2}{3}{sin}^{-1}\left\{{\left(\frac{x}{a}\right)}^{\frac{3}{2}}\right\}+c$$$$check$$$$y=\frac{2}{3}{sin}^{-1}\left\{{\left(\frac{x}{a}\right)}^{\frac{3}{2}}\right\}+c$$$$\frac{dy}{dx}=\frac{2}{3}\times \frac{1}{\sqrt{1-{\left(\frac{x}{a}\right)}^3}}\times \frac{3}{2}\times {\left(\frac{x}{a}\right)}^{\frac{1}{2}}\times \frac{1}{a}$$$$=a^{\frac{3}{2}-1-\frac{1}{2}}.\frac{\sqrt{x}}{\sqrt{a^3-x^3}}=\frac{\sqrt{x}}{\sqrt{a^3-x^3}}$$$$$$

Answered by MJS last updated on 27/Jul/18

$$a\ne 0$$$$\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}}dx=$$$$[t=\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}3\to dx=\frac{2a^{\frac{3}{2}}}{3\sqrt{x}}dt]$$$$=\frac{2}{3}\int \frac{1}{\sqrt{1-t^2}}=\frac{2}{3}\arcsin t=\frac{2}{3}\arcsin {\left(\frac{x}{a}\right)}^{\frac{3}{2}}+C$$$$$$$$a=0$$$$\int \frac{\sqrt{x}}{\sqrt{-x^3}}dx=\int \sqrt{-\frac{1}{x^2}}dx=\mathrm{i}\int \frac{dx}{\mid x\mid }=\mathrm{i}\times \mathrm{sign}\left(x\right)\ln \mid x\mid +C$$

Answered by math khazana by abdo last updated on 27/Jul/18

$$letI=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}}dxchangementx=atgive$$$$I=\int \frac{\sqrt{a}\sqrt{t}}{\sqrt{a^3}\sqrt{1-t^3}}adt=\int \frac{\sqrt{t}}{\sqrt{1-t^3}}dt$$$${=}_{t^3={sin}^2\alpha }\int \frac{\sqrt{sin{\alpha }^{\frac{2}{3}}}}{cos\alpha }\frac{2}{3}cos\alpha sin{\alpha }^{\frac{2}{3}-1}d\alpha (t=sin{\alpha }^{\frac{2}{3}})$$$$=\frac{2}{3}\int \frac{{\left(sin\alpha \right)}^{\frac{1}{3}}{\left(sin\alpha \right)}^{-\frac{1}{3}}}{cos\alpha }cos\alpha d\alpha$$$$=\frac{2}{3}\int d\alpha =\frac{2}{3}\alpha +c=\frac{2}{3}arcsin\left(t^{\frac{3}{2}}\right)+c$$$$=\frac{2}{3}arcsin\left({\left(\frac{x}{a}\right)}^{\frac{3}{2}}\right)+c\Rightarrow$$$$I=\frac{2}{3}arcsin\left(\sqrt{\frac{x^3}{a^3}}\right)+c(wesupposea>0)$$

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