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

Toolbox:
• $\int udv=uv-\int vdu$
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,
$I=\int\sqrt{a^2-x^2}dx$
This is of the form $\int udv=uv-\int vdu$
Let $u=\sqrt{a^2-x^2}$
$du=\large\frac{1}{2}$$(a^2-x^2)^{-\large\frac{1}{2}}(-2x)dx \;\;=-\large\frac{x}{\sqrt{a^2-x^2}}$$dx$
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)$$dx \;\;=x\sqrt{a^2-x^2}-\int\large\frac{-x^2}{\sqrt{a^2-x^2}}dx Consider I_1=-\int\large\frac{x^2}{\sqrt{a^2-x^2}}$$dx$
Add and subtract $a^2$
$I_1=\int\large\frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}$$dx On separating we get, I_1=\int\sqrt{a^2-x^2}dx+a^2\int\large\frac{1}{\sqrt{a^2-x^2}}$$dx$
But $\int\sqrt{a^2-x^2}dx=I$
Substituting we get,
$I_1=I+a^2\int\large\frac{1}{\sqrt{a^2-x^2}}$$dx Step 3: I=x\sqrt{a^2-x^2}-I+a^2\int\large\frac{1}{\sqrt{a^2-x^2}}$$dx$
$2I=x\sqrt{a^2-x^2}+a^2\int\large\frac{dx}{\sqrt{a^2-x^2}}$
$2I=x\sqrt{a^2-x^2}+a^2\sin^{-1}\big(\large\frac{x}{a}\big)+c$
$I=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)+c$
Hence proved.