logo

Ask Questions, Get Answers

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

Integrate the function\[\sqrt{1-4x-x^2}\]

$\begin{array}{1 1} \frac{(x+2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \sin^{-1}( \frac{x+2}{\sqrt 5})+c \\ \frac{(x+2) (\sqrt{1-4x-x^2})}{2}-\frac{5}{2} \cos^{-1}( \frac{x+2}{\sqrt 5})+c \\ \frac{(x-2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \sin^{-1}( \frac{x-2}{\sqrt 5})+c \\ \frac{(x-2) (\sqrt{1-4x-x^2})}{2}+\frac{5}{2} \cos^{-1}(\frac{x-2}{\sqrt 5})+c \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac {a^2}{2}\sin^{-1}(\frac{x}{a})+c$
Given $I=\sqrt{1-4x-x^2}dx$
 
$(1-4x-x^2)=-(x^2+4x-1)$
 
This can split as $-[(x^2+4x+4)-4-1]$
 
                        $=-[(x+2)^2-(\sqrt 5)^2]$
 
                        $I=[(\sqrt 5)^2-(x+2)^2]$
 
           Hence $I=\int \sqrt{(\sqrt 5)^2-(x+2)^2}dx$
 
This is of the form $\sqrt{a^2-x^2} dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}(\frac{x}{a})+c$
 
                        Here $a=\sqrt 5.x=(x+2)$
       
Hence$\int \sqrt{1-4x-x^2}=\frac{(x+2)}{2}\sqrt{1-4x-x^2}+\frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt 5})+c$

 

answered Feb 7, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...