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Evaluate the following $\int\sqrt {5-2x+x^2}dx$

$\begin{array}{1 1} (A)\; \large\frac{x+1}{6} \sqrt {5-2x-x^2}+ 3 \sin ^{-1}\bigg(\frac{x+1}{\sqrt 6}\bigg)+c \\(B)\;\large\frac{x+1}{2} \sqrt {5-2x-x^2}+ 6 \sin ^{-1}\bigg(\frac{x+1}{\sqrt 6}\bigg)+c \\ (C)\;\large\frac{x+1}{2} \sqrt {5-2x-x^2}- 3 \sin ^{-1}\bigg(\frac{x+1}{\sqrt 6}\bigg)+c \\ (D)\;\large\frac{x+1}{2} \sqrt {5-2x-x^2}+ 3 \sin ^{-1}\bigg(\frac{x+1}{\sqrt 6}\bigg)+c \end{array} $

1 Answer

  • $\int \sqrt {a^2-x^2}dx= \large\frac{x}{2} \sqrt {a^2-x^2}+ \frac{a^2}{2} \sin ^{-1} \bigg(\frac{x}{a}\bigg)+c$
Let I =$\int\sqrt {5-2x+x^2}dx$
consider $(5-2x+x^2)$
By completing the squares, this can be written as
$=-\bigg[(x+1)^2-(\sqrt 6)^2\bigg]=(\sqrt 6)^2-(x+1)^2$
Hence $I=\sqrt {(\sqrt 6)^2-(x+1)^2}dx$
This is of the form $\sqrt {a^2-x^2}dx$
$=\large\frac{x}{2} \sqrt {a^2-x^2}+\frac{a^2}{2} \sin ^{-1}(\frac{x}{a})+c$
Here $x=(x+1)\; and\; a =\sqrt 6$
Hence on integrating $\int \sqrt {5-2x-x^2}dx$
$=\large\frac{x+1}{2} \sqrt {5-2x-x^2}+\frac {(\sqrt 6)^2}{2} \sin ^{-1}\bigg( \frac{x+1}{\sqrt 6}\bigg)+c$
$=\large\frac{x+1}{2} \sqrt {5-2x-x^2}+ 3 \sin ^{-1}\bigg(\frac{x+1}{\sqrt 6}\bigg)+c$
answered Apr 16, 2013 by meena.p