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Integrate the function $\int \frac{1}{\cos^2x(1-\tan x)^2}$

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

1 Answer

  • (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,
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