Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Integrate the function\[\sqrt{1+3x-x^2}\]

$\begin{array}{1 1}\large \frac{(2x-3)(\sqrt{1+3x-x^2})}{4} +\frac{13 }{8}\sin^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\\large \frac{(2x-3)(\sqrt{1+3x-x^2})}{4} -\frac{13 }{8}\cos^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\ \large \frac{(3x+2)(\sqrt{1+3x-x^2})}{4} +\frac{13 }{8}\cos^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\ \large \frac{(3-2x)(\sqrt{1+3x-x^2})}{4} -\frac{13 }{8}\sin^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \end{array} $

Can you answer this question?

1 Answer

0 votes
  • $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac {a^2}{2}\sin^{-1}(\frac{x}{a})+c$
Given $I=\sqrt{1+3x-x^2}dx$
Consider $1+3x-x^2=-(x^2-3x-1)$
This can split as $-[(x^2-3x+\frac{9}{4})-\frac{9}{4}-1]$
                         $=-[(x-\frac{3}{2})^2-\frac {9}{4}-1]$
                         $=-[(x-\frac{3}{2})^2-(\frac {\sqrt {13}}{2})^2]$
                          $I=[(\frac{\sqrt {13}}{2})^2-(x-\frac{3}{2})^2]$
         Hence $I=\int \sqrt{(\frac{\sqrt {13}}{2})^2-(x-\frac{3}{2})^2} dx$
Clearly This is of the form $\int \sqrt{a^2-x^2} dx$
Here in the place a we have $\frac{\sqrt{13}}{2}$ and in the place of x we have (x-3/2).
Therefore $\int \sqrt{1+3x-x^2}dx=\frac{(x-\frac{3}{2})}{2}\sqrt{1+3x-x^2}+\frac{(\frac{\sqrt {13}}{2})^2}{2}\sin^{-1}(\frac{x-3/2}{\frac{\sqrt {13}}{2}})+c$
On simplifying we get,
$\frac{2x-3}{4} \sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}(\frac{2x-3}{\sqrt {13}})+c$
answered Feb 7, 2013 by meena.p
edited Apr 11, 2016 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App