Question #80cb5

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Question #80cb5

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Answer 1 · 2017-04-08T15:04:29

$\frac{2 \sqrt{\frac{x}{a^{3} - x^{3}}} \sqrt{a^{3} - x^{3}} {tan}^{- 1} (\frac{x^{\frac{3}{2}}}{\sqrt{a^{3} - x^{3}}})}{3 \sqrt{x}} + C$ $(2 sqrt(x/(a^3 - x^3)) sqrt(a^3 - x^3) tan^(-1)(x^(3/2)/sqrt(a^3 - x^3)))/(3 sqrt(x)) + C$

Explanation:

Simplify powers

$\int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x$ $int sqrt(x) / sqrt(a^3 - x^3) dx$

Substitute u = $x^{\frac{3}{2}}$ $x^(3/2)$ and du = $\frac{3 \sqrt{x}}{2}$ $(3sqrt(x)) / 2$

$= \frac{2}{3} \int \frac{1}{\sqrt{a^{3} - u^{2}}} d u$ $= 2/3 int1/sqrt(a^3 - u^2) du$

Take a^3 out of the root

$\frac{2}{3} \int \frac{1}{a^{\frac{3}{2}} \sqrt{1 - \frac{u^{2}}{a^{3}}}} d u$ $2/3 int1/(a^(3/2) sqrt(1 - u^2/a^3)) du$

Factor out contsants

$\frac{2}{3 a^{\frac{3}{2}}} \int \frac{1}{\sqrt{1 - \frac{u^{2}}{a^{3}}}} d u$ $2/(3 a^(3/2)) int1/sqrt(1 - u^2/a^3) du$

Substitute r = $\frac{u}{a^{\frac{3}{2}}}$ $u / a^(3/2)$ and dr = $a^{- \frac{3}{2}}$ $a^(-3/2)$ du

$\frac{2}{3} \int \frac{1}{\sqrt{1 - r^{2}}} d r$ $2/3 int1/sqrt(1 - r^2) dr$

The integral of $\frac{1}{\sqrt{1 - r^{2}}} i s {sin}^{- 1} (r)$ $1/sqrt(1 - r^2) is sin^(-1)(r)$

$= \frac{2}{3} {sin}^{- 1} (s) + C$ $= 2/3 sin^(-1)(s) + C$

Substitute back r and u

$= \frac{2}{3} {sin}^{- 1} (\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}) + C$ $= 2/3 sin^(-1)(x^(3/2)/a^(3/2)) + C$

Remove negative exponents

$\frac{2 \sqrt{\frac{x}{a^{3} - x^{3}}} \sqrt{a^{3} - x^{3}} {tan}^{- 1} (\frac{x^{\frac{3}{2}}}{\sqrt{a^{3} - x^{3}}})}{3 \sqrt{x}} + C$ $(2 sqrt(x/(a^3 - x^3)) sqrt(a^3 - x^3) tan^(-1)(x^(3/2)/sqrt(a^3 - x^3)))/(3 sqrt(x)) + C$

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Question #80cb5

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