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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{\cos2x}{(\cos x+\sin x)^2}\]

$\begin{array}{1 1}\log | \sin x+\cos x |+c. \\ \log | \sin x-\cos x |+c \\ \log | 2\sin x+2\cos x |+c \\ \log | 2\sin x-\cos x |+c. \end{array} $

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1 Answer

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Toolbox:
  • $(i)\;\sin2x=2\sin x\cos x.$
  • $(ii)\;\int\frac{1}{x}dx=log x+c.$
  • $(iii)\;Method\;of\;substitution:$
  • If f(x)=t.Therefore f'(x)dx=dt.Thus $\int f(x)dx=\int t.dt.$
Given:$I=\int\frac{\cos 2x}{(\cos x+\sin x)^2}dx.$
 
$\;\;\;=\int\frac{\cos 2x}{\cos^2+\sin^2x+2\sin x\cos x}dx.$
 
But $2\sin x\cos x=\sin 2x$ and $\sin^2x+\cos^2x=1.$
 
$\;\;\;=\int\frac{\cos 2x}{1+\sin2x}dx.$
 
Let $1+\sin 2x=t.$
 
On differentiating
 
$2\cos 2xdx=dt.$
 
On substituting t we get,
 
$I=\int\frac{dt/2}{t}=\frac{1}{2}\int\frac{1}{t}.dt.$
 
$\sec^2xdx=dt.$
 
By substituting t we get,
 
$I=\int t.dt+\int\frac{1}{t}.dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}log |t|+c.$
 
On Substituting back for t we get,
 
$\;\;\;=\frac{1}{2}log|1+\sin 2x|+c.$
 
But $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$.
 
$\;\;\;=\frac{1}{2}log|\sin^2x+\cos^2x+2\sin x\cos x|+c.$
 
This is of the form $(a+b)^2$
 
$\;\;\;=\frac{1}{2}log(\sin x+\cos x)^2+c.$
 
$\frac{1}{2}log(\sin x+\cos x)^2$can be written as $log[(\sin x\tan x)^2]$
 
$\;\;\;=log\sqrt {(\sin x+\cos x)^2}+c.$
 
$\;\;\;=log|\sin x+\cos x|+c.$

 

answered Feb 3, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans
 
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