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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\int \frac{1}{\cos^2x(1-\tan x)^2}$

$\begin{array}{1 1}\large \frac{1}{(1-\tan x)}+c \\\large \frac{1}{(1+\tan x)}+c\\\large \frac{1}{(1-\cos x)}+c \\ \large \frac{1}{(1-\sin x)}+c \end{array}$

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1 Answer

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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)$\int \tan xdx=\sec^2x+c.$
Given $I=\int \frac{1}{\cos^2x(1-\tan x)^2}dx.$
 
$\;\;\;\;\qquad=\int \frac{\sec^2xdx}{(1-\tan x)^2}$
 
Put $1-\tan x=t.$
 
$-\sec^2xdx=dt\Rightarrow \sec^2xdx=-dt.$
 
Now substituting for x and dx we get,
 
$I=\int\frac{-dt}{t^2}=-\int t^{-2}dt .$
 
On integrating we get,
 
$-\begin{bmatrix}\frac{t^{-1}}{-1}\end{bmatrix}+c=\frac{1}{t}+c.$
 
Substituting back for t we get,
 
$\int\frac{1}{\cos^2x(1-\tan x)^2}=\frac{1}{(1-\tan x)}+c.$
 
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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