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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx$. Substitute $x = a(\sin 	heta)^{2/3}$. Then $dx = a \cdot \frac{2}{3}(\sin 	heta)^{-1/3} \cdot \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \sqrt{\frac{x}{a^3 - x^3}} \, dx$. Substitute $x = a(\sin 	heta)^{2/3}$. Then $dx = a \cdot \frac{2}{3}(\sin 	heta)^{-1/3} \cdot \cos 	heta \, d	heta$. Substituting these into the integral: $I = \int \sqrt{\frac{a(\sin 	heta)^{2/3}}{a^3 - a^3(\sin 	heta)^2}} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \int \sqrt{\frac{a(\sin 	heta)^{2/3}}{a^3(1 - \sin^2 	heta)}} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \int \frac{a^{1/2}(\sin 	heta)^{1/3}}{a^{3/2} \cos 	heta} \cdot \frac{2}{3}a(\sin 	heta)^{-1/3} \cos 	heta \, d	heta$ $I = \frac{2}{3} \int d	heta = \frac{2}{3} 	heta + c$. Since $x = a(\sin 	heta)^{2/3}$,we have $(\frac{x}{a})^{3/2} = \sin 	heta$,so $	heta = \sin^{-1}(\frac{x}{a})^{3/2}$. Thus,$I = \frac{2}{3} \sin^{-1}(\frac{x}{a})^{3/2} + c$.

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

Similar Questions

$\int \frac{x^2}{(\sqrt{4-x^2})^3} dx =$

$\int \sin^5 x \, dx =$

$\int {\frac{{{e^{{{\tan }^{ - 1}}\sqrt x }}}}{{\sqrt x + x\sqrt x }}} dx = $

If $\int \frac{\log _e(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx = f(g(x)) + c$, then:

If $\int \sqrt{\frac{x}{a^3-x^3}} d x=g(x)+c$,then $g(x)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module