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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large e^{2x}-e^{-2x}}{\large e^{2x}+e^{-2x}}$

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Toolbox:
  • (i)Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
  • (ii)$\int e^{ax}dx=\frac{1}{a}e^{ax}+c.$
Given $I=\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx $.
 
Let $e^{2x}+e^{-2x}=t.$
 
$(2e^2x-2e^{-2x})dx=dt.$
 
$\Rightarrow 2(e^{2x}-e^{-2x})dx=dt.\Rightarrow (e^{2x}-e^{-2x})dx=\frac{dt}{2}.$
 
Now substituting for x and dx we get,
 
$I=\int\frac{dt}{2t}=\frac{1}{2}\int\frac{dt}{t}.$
 
On integrating we get,
 
$\frac{1}{2}log|t|+c.$
 
Substituting back for t we get,
 
$\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx=\frac{1}{2}log|e^{2x}+e^{-2x}|+c.$
 
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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