If int frac{x^2}{sqrt{1-x}} ,d x = p sqrt{1-x | vedclass

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(A) Let $I = \int \frac{x^2}{\sqrt{1-x}} \,dx$. Substitute $1-x = t$, so $x = 1-t$ and $dx = -dt$. $I = \int \frac{(1-t)^2}{\sqrt{t}} (-dt) = -\int \f

Vedclass · Accepted Answer

$\frac{-2}{15}$

Vedclass · Answer

(A) Let $I = \int \frac{x^2}{\sqrt{1-x}} \,dx$. Substitute $1-x = t$, so $x = 1-t$ and $dx = -dt$. $I = \int \frac{(1-t)^2}{\sqrt{t}} (-dt) = -\int \frac{1-2t+t^2}{t^{1/2}} dt$. $I = -\int (t^{-1/2} - 2t^{1/2} + t^{3/2}) dt$. $I = -(2t^{1/2} - \frac{4}{3}t^{3/2} + \frac{2}{5}t^{5/2}) + c$. $I = -\frac{2}{15} t^{1/2} (15 - 10t + 3t^2) + c$. Since $t = 1-x$, $I = -\frac{2}{15} \sqrt{1-x} (15 - 10(1-x) + 3(1-x)^2) + c$. $I = -\frac{2}{15} \sqrt{1-x} (15 - 10 + 10x + 3(1 - 2x + x^2)) + c$. $I = -\frac{2}{15} \sqrt{1-x} (5 + 10x + 3 - 6x + 3x^2) + c$. $I = -\frac{2}{15} \sqrt{1-x} (3x^2 + 4x + 8) + c$. Comparing this with $p \sqrt{1-x} (3x^2 + 4x + 8) + c$, we get $p = \frac{-2}{15}$.

If $\int \frac{x^2}{\sqrt{1-x}} \,d x = p \sqrt{1-x} (3x^2 + 4x + 8) + c$ where $c$ is a constant of integration, then the value of $p$ is

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