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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{1}{\sin x\cos^3x}\]

$\begin{array}{1 1} \frac{\tan^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}-\log|\tan x|+c \\ \frac{\cot^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}+\log|\cot x|+c \end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $(i)\;\sin^2x+\cos^2x=1.$
  • $(ii)\;Method\;of\;substitution.$
  • Let f(x)=t.
  • Therefore f'(x)dx=dt.Thus $\int f(x)dx=\int t.dt.$
Given:$I=\int\frac{1}{\sin x\cos x}dx.$
 
This can be written as,
 
$I=\frac{\sin^2x+\cos^2x}{\sin x\cos^3x}dx.$
 
Separating the terms we get,
 
$I=\int\frac{\sin^2x}{\sin x\cos^3x}dx+\int\frac{\cos^2x}{\sin x\cos^3x}dx.$
 
$\;\;\;=\int\frac{\sin x}{\cos x}\frac{1}{\cos^2x}dx+\int\frac{\cos x}{\sin x}\frac{1}{\cos^2x}dx.$
 
$\;\;\;=\int\tan x.\sec xdx+\int\frac{\sec^2}{\tan x}dx.$
 
Let tan x=t.
 
On differentiating we get,
 
$\sec^2xdx=dt.$
 
By substituting t we get,
 
$I=\int t.dt+\int\frac{1}{t}.dt.$
 
On integrating we get,
 
$\;\;\;=\frac{t^2}{2}+log |t|+c.$
 
Substituting for t we get,
 
$\int\frac{\sin^2x+\cos ^2x}{\sin x\cos^3x}dx=\frac{\tan^2x}{2}+log|\tan x|+c.$

 

answered Feb 3, 2013 by sreemathi.v
 
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