int frac{3x^2}{sqrt{9 - 16x^6}} , dx = | vedclass

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(A) Let $I = \int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx$. We can rewrite the integral as: $I = \int \frac{3x^2}{\sqrt{3^2 - (4x^3)^2}} \, dx$. Let $t =

Vedclass · Accepted Answer

$\frac{1}{4} \sin^{-1} \left( \frac{4x^3}{3} \right) + c$

Vedclass · Answer

(A) Let $I = \int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx$. We can rewrite the integral as: $I = \int \frac{3x^2}{\sqrt{3^2 - (4x^3)^2}} \, dx$. Let $t = 4x^3$. Then $dt = 12x^2 \, dx$,which implies $3x^2 \, dx = \frac{1}{4} dt$. Substituting these into the integral: $I = \frac{1}{4} \int \frac{dt}{\sqrt{3^2 - t^2}}$. Using the standard formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + c$: $I = \frac{1}{4} \sin^{-1} \left( \frac{t}{3} \right) + c$. Substituting $t = 4x^3$ back: $I = \frac{1}{4} \sin^{-1} \left( \frac{4x^3}{3} \right) + c$.

$\int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx = $

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