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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\tan^2(2x-3)$

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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)$\int \sec^2x=\tan x+c.$
Given $I=\int \tan^2(2x-3)dx.$
 
Put t=2x-3.
 
dt=2dx $\Rightarrow dx=\frac{dt}{2}.$
 
Now substituting for x and dx we get,
 
$I=\int\tan^2t,\frac{dt}{2}=\frac{1}{2}\int\tan^2t.$
 
But $\tan^2t=(\sec^2t-1)$
 
$I=\frac{1}{2}\int(\sec^2t-1)dt.$
 
$\;\;\;=\frac{1}{2}[\int\sec^2t-\int dt].$
On integrating we get,
 
$\frac{1}{2}[\tan t-t]+c.$
 
Substituting back for t we get,
 
$\;\;\;=\frac{1}{2}[\tan(2x-3)-(2x-3)]+c.$
 
$\;\;\;=\frac{1}{2}[\tan(2x-3)]-x+c.(\frac{3}{2}$ is a constant)
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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