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Prove that : $\large \frac{d}{dx} \bigg[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}\bigg( \frac{x}{a} \bigg) \bigg] = \sqrt{a^2-x^2} $

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  • $\int udv=uv-\int vdu$
Step 1:
Given :
$\large \frac{d}{dx} \bigg[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}\bigg( \frac{x}{a} \bigg) \bigg] = \sqrt{a^2-x^2} $
(i.e) $\int \sqrt{(a^2-x^2)dx}=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$
To prove the above statement let,
This is of the form $\int udv=uv-\int vdu$
Let $u=\sqrt{a^2-x^2}$
Step 2:
$dv=dx\Rightarrow v=x$
Now substituting for $u,v,du$ and $dv$ we get,
$I=x\sqrt{a^2-x^2}-\int x.\big(-\large\frac{x}{\sqrt{a^2-x^2}}\big)$$dx$
Consider $I_1=-\int\large\frac{x^2}{\sqrt{a^2-x^2}}$$dx$
Add and subtract $a^2$
On separating we get,
But $\int\sqrt{a^2-x^2}dx=I$
Substituting we get,
Step 3:
Hence proved.
answered Nov 14, 2013 by sreemathi.v

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