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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int\frac{\sqrt x}{\sqrt{a^3-x^3}}dx$

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Toolbox:
  • $\large\int \frac{dx}{\sqrt {a^2-x^2}}=\sin ^{-1}(\frac{x}{a})+c$
  • If $f(x)$ is substituted by $f(t),$ then $f'(x) dx=f'(t)dx$
  • Hence $\int f(x) dx=\int f(t)dt$
$\large\int\frac{\sqrt x}{\sqrt{a^3-x^3}}dx$
This can be written as $\large\int \frac{\sqrt x}{\sqrt {(a^{3/2})^2-(x^{3/2})^2}}dx$
Put $x^{3/2}=t$. on differentiating with respect to t
we get $\frac{3}{2} x^{1/2} dx=dt =>dx=\frac{2}{3 \sqrt x}dt$
On substituting this we get
Therefore $I=\Large\int \frac{\frac{2}{3 \sqrt x} \times \sqrt x.dt}{\sqrt { (a^{3/2})^2-t^2}}$
$=\large\frac{2}{3} \int \frac{dt}{\sqrt {(a^{3/2})^2-t^2}}$
This is of the form $\large\frac{dx}{\sqrt {a^2-x^2}}$
$=\sin^{-1}(x/a)$
Here $x=t\;and \;a=a^{3/2}$
Therefore on integrating I we get
$I=\large\frac{2}{3} \sin ^{-1} \bigg(\frac{a^{3/2}}{t}\bigg)+c$
Substituting for t we get,
$I=\large\frac{2}{3} \sin ^{-1} \bigg(\frac{a^{3/2}}{x^{3/2}}\bigg)+c$
answered Apr 16, 2013 by meena.p
 
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