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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\tan^4x\]

$\begin{array}{1 1} \frac{1}{3} \tan^3x-\tan x+x+c \\ \frac{1}{3} \tan^3x+\tan x+x+c. \\ \frac{1}{3} \tan^3x-\tan x-x+c. \\ \frac{1}{3} \tan^3x+\tan x-x+c.\end{array} $

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1 Answer

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Toolbox:
  • $(i)\tan^2x=\sec^2x-1$
  • $(ii)\int \sec^2xdx=\tan x+c.$
Given $I=\int\tan^4xdx.$
 
$\tan^4x=\tan^2x.\tan^2xdx$ .
 
$\;\;\;=\int \tan^2x,\tan^2xdx.$
 
This can be written as,
 
$\;\;\;=\int(\sec^2x-1).\tan^2xdx.$
 
On expanding we get,
 
$\int(\sec^2x.\tan^2x-\tan^2x)dx.$
 
On separating the term,
 
$\int\sec^2x.\tan^2xdx-\int(\sec^2x-1)dx.$
 
$I=\int\sec^2x\tan^2xdx-\int\sec^2x-\int dx.$
 
Let $\tan x=t$.
 
On differentiating
 
$\sec^2xdx=dt.$
 
Substituting this we get,
 
$I=\int t^2.dt-\sec^2xdx-\int dx.$
 
On integrating we get:
 
$\frac{t^3}{3}-\tan x+x+c.$
 
Substituting for t we get,
 
$\frac{(\tan x)^3}{3}-\tan x+x+c.$
 
$\;\;\;=\frac{1}{3}\tan^3x-\tan x+x+c.$
 

 

answered Jan 31, 2013 by sreemathi.v
 
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