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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{1}{1+\cot x}\]

$\begin{array}{1 1} \frac{x}{2}-\frac{1}{2} \log(\sin x+\cos x)+c \\ \frac{x}{2}+\frac{1}{2} \log(\sin x+\cos x)+c \\ \frac{x}{2}-\frac{1}{2} \log(\sin x-\cos x)+c \\ \frac{x}{2}+\frac{1}{2} \log(\sin x-\cos x)+c\end{array}$

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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 $I=\int \frac{1}{1+\cot x}dx=\int\frac{1}{1+\frac{\cos x}{\sin x}}dx=\int\frac{\sin x}{\sin x+\cos x}.$
 
Multiply and divide by 2
 
$I=\frac{1}{2}\int\frac{2\sin x}{\sin x+\cos x}dx.$
 
Add and subtract cos x in the numerator
 
$I=\frac{1}{2}\int\frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x}dx.$
 
Now separating the terms we get,
 
$I=\frac{1}{2}\int dx+\frac{1}{2}\int\frac{(\sin x-\cos x)}{\sin x+\cos x}dx.$
 
$\;\;\;=\frac{1}{2}(x)+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x+\cos x}dx.$
Put $\sin x+\cos x=t.$
 
$(\cos x-\sin x)dx=dt.$
 
$-(\sin x-\cos x)dx=dt.$
 
$(\sin x-\cos x)dx=-dt.$
 
Substituting for x and dx we get,
 
$I=\frac{x}{2}+\frac{1}{2}\int\frac{-dt}{t}.$
 
On integrating we get,
 
$I=\frac{x}{2}-\frac{1}{2}(log t)+c.$
 
Substituting back for t we get,
 
$I=\frac{x}{2}-\frac{1}{2}log(\sin x+\cos x)+c.$

 

answered Jan 29, 2013 by sreemathi.v
 
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