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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{x+2}{\sqrt{4x-x^2}}\]

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

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Toolbox:
  • $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,where p,q,a,b,c are constants,we are to find real numbers such that \[px+q=A\frac{d}{dx}(ax^2+bx+c)+B.\]
  • Therefore $px+q=A(2ax+b)+B.$
  • To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms:
  • (ii)$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+c.$
Given:$I=\int \frac{x+2}{\sqrt{4x-x^2}}dx.$
 
Let x+2=$A\frac{d}{dx}(4x-x^2)+B.$
 
$\;\;x+2=A(4-2x)+B.$
 
Now equating the coefficients of x we get,
 
1=-2A$\Rightarrow A=\frac{-1}{2}.$
 
equating the constants,
 
2=4A+B$\Rightarrow B=4.$
 
Hence A=-1/2 and B=4.
 
$I=\int\frac{\frac{-1}{2}(4-2x)+4}{\sqrt{4x-x^2}}dx.$
 
On separating the terms we can write,
 
$I=\frac{-1}{2}\int\frac{a-2x}{\sqrt{4x-x^2}}+4\int\frac{dx}{\sqrt{4x-x^2}}.$
 
This can be written as,
 
$I=\frac{-1}{2}\int\frac{dt}{\sqrt t}+4\int\frac{dx}{\sqrt{4x-x^2}}.$
 
$(2x-9)dx=dt.$
 
$4x-x^2=-(x^2-4x).$
 
$\;\;\;=-[(x-2)^2-(2)^2]$
 
$\;\;\;=2^2-(x-2)^2$
 
$I=\frac{-1}{2}\int\frac{dt}{\sqrt t}+4\int\frac{dx}{\sqrt{2^2-(x-2)^2}}.$
 
On integrating we get,
 
$\;\;\;=\frac{-1}{2}\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+4\sin^{-1}\big(\frac{x-2}{2}\big)+c.$
 
$\;\;\;=\frac{-1}{2}\big(\frac{\sqrt t}{1/2}\big)+4\sin^{-1}\frac{x-2}{2}+c.$
 
Substituting for t we get,
 
$\;\;\;=\frac{-1}{2}\big(\frac{\sqrt {4x-x^2}}{1/2}\big)+4\sin^{-1}\frac{x-2}{2}+c.$
 
Hence $\int\frac{x+2}{\sqrt{4x-x^2}}dx=4\sin^{-1}\big(\frac{x-2}{2}\big)-\sqrt{4x-x^2}+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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