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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{1}{1-\ tanx}\]

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

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1 Answer

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Toolbox:
  • 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 $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{1}{1+\tan x}dx=\int\frac{1}{1+\frac{\sin x}{\cos x}}dx.$
 
$I=\int\frac{\cos x}{\sin x+\cos x}dx.$
 
Multiply and divide by 2
 
$I=\int\frac{2\cos x}{\sin x+\cos x}dx.$
 
Add and subtract sinx in the numerator,
 
$I=\int\frac{(\sin x+\cos x)+(\cos x-\sin x)}{\sin x+\cos x}dx$.
 
separating terms we get,
 
$I=\int dx+\int\frac{(\cos x-\sin x)}{\sin x+\cos x}dx.$
 
Put $\sin x+\cos x=t.$
 
$(\cos x-\sin x)dx=dt.$
 
$I=\int dx+\int\frac{dt}{t}.$
 
On integrating we get,
 
x+log|t|+c.
 
Substituting back for t we get,
 
$\int\frac{1}{1+\tan x}dx=x+log(\sin x+\cos x)+c.$

 

answered Jan 29, 2013 by sreemathi.v
 
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