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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int\sqrt {2ax-x^2}dx$

$\begin{array}{1 1} (A)\;\large\frac{x-a}{2} \sqrt {2ax-x^2}-\frac{a^2}{2} \sin ^{-1} \bigg(\frac{x-a}{a}\bigg)+c \\ (B)\;\large\frac{x-a}{4} \sqrt {2ax-x^2}+\frac{a^2}{2} \sin ^{-1} \bigg(\frac{x-a}{a}\bigg)+c \\ (C)\large\frac{x-a}{2} \sqrt {2ax-x^2}+\frac{a^2}{2} \sin ^{-1} \bigg(\frac{x-a}{a}\bigg)+c \\(D)\;\large\frac{x-a}{4} \sqrt {2ax-x^2}-\frac{a^2}{2} \sin ^{-1} \bigg(\frac{x-a}{a}\bigg)+c\end{array} $

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  • $\large\int \sqrt {a^2-x^2}= \large\frac{x}{2} \sqrt {a^2-x^2}+ \frac{a^2}{2} \sin ^{-1} \bigg(\frac{x}{a}\bigg)+c$
$\int \sqrt {2ax-x^2} dx$
Consider $(2ax-x^2)$
=$-(x^2-2ax)$
By completing the squares we can write this as $-(x-a)^2-a^2$
$=(a^2)-(x-a)^2$
Therefore $I=\int \sqrt {a^2-(x-a)^2} dx$
This is of the form: $\large\int \sqrt {a^2-x^2}=\frac{x}{2} \sqrt {a^2-x^2}+\frac{a^2}{2} \sin ^{-1} (\frac{x}{a})+c$
Here $x=x-2a\; and\; a=2a$
Now on integrating we get
$\large\int \sqrt {a^2-(x-a)^2} dx =\frac{x-a}{2} \sqrt {2ax-x^2}+\frac{a^2}{2} \sin^{-1} \bigg(\frac{x-a}{a} \bigg)+c$
$=\large\frac{x-a}{2} \sqrt {2ax-x^2}+\frac{a^2}{2} \sin ^{-1} \bigg(\frac{x-a}{a}\bigg)+c$
answered Apr 16, 2013 by meena.p
 
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