int sqrt{frac{x}{a^3 - x^3}} , dx = | vedclass

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Question

(B) ધારો કે $I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx$. $x = a(\sin 	heta)^{2/3}$ આદેશ લેતા. તેથી $dx = a \cdot \frac{2}{3}(\sin 	heta)^{-1/3} \cdo

Vedclass · Accepted Answer

$\frac{2}{3}\sin^{-1}\left(\frac{x}{a}\right)^{3/2} + c$

Vedclass · Answer

(B) ધારો કે $I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx$. $x = a(\sin 	heta)^{2/3}$ આદેશ લેતા. તેથી $dx = a \cdot \frac{2}{3}(\sin 	heta)^{-1/3} \cdot \cos 	heta \, d	heta$. આ કિંમતો સંકલનમાં મૂકતા: $I = \int \sqrt{\frac{a(\sin 	heta)^{2/3}}{a^3 - a^3(\sin 	heta)^2}} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \int \sqrt{\frac{a(\sin 	heta)^{2/3}}{a^3(1 - \sin^2 	heta)}} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \int \frac{a^{1/2}(\sin 	heta)^{1/3}}{a^{3/2} \cos 	heta} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \frac{2}{3} \int d	heta = \frac{2}{3} 	heta + c$. કારણ કે $x = a(\sin 	heta)^{2/3}$,તેથી $(\frac{x}{a})^{3/2} = \sin 	heta$,એટલે કે $	heta = \sin^{-1}(\frac{x}{a})^{3/2}$. આમ,$I = \frac{2}{3} \sin^{-1}(\frac{x}{a})^{3/2} + c$.

$\int \sqrt{\frac{x}{a^3 - x^3}} \, dx = $

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