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Home  >>  CBSE XII  >>  Math  >>  Integrals

Find the integrals of the functions\[\tan^32x\sec2x\]

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Toolbox:
  • $\tan^2x=\sec^2x-1.$
  • $(ii)\int \sec^2xdx=\tan x+c.$
Given $I=\int\tan^32x\sec2xdx.$
 
$\tan^32x$ can be written as $\tan^22x\tan 2x$.
 
$\;\;\;=\int\tan^22x.\tan2x.\sec 2xdx.$
 
But $\tan^22x=\sec^22x-1.$
 
$\;\;\;=\int[(\sec^22x-).\tan2x.\sec 2x]dx.$
 
On multiplying,
 
$\;\;\;=\int\sec^22x.\tan2x.\sec 2x-\int\sec2x.\tan 2xdx.$
 
Put $\sec2x=t.$
 
On differentiating
 
$2\sec2x\tan 2xdx=dt.$
 
$(\sec2x\tan 2x)dx=\frac{dt}{2}.$
 
On substituting we get,
 
$I=\frac{1}{2}\int t^2 dt-\int\sec2x\tan 2xdx.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}[\frac{t^3}{3}]-\frac{\sec 2x}{2}+c.$
 
Substituting for t we get,
 
$\;\;\;=-\frac{1}{2}(\frac{\sec2x}{3})^3-\frac{\sec2x}{2}+c.$
 
$\;\;\;=\frac{(\sec2x)^3}{6}-\frac{\sec2x}{2}+c.$
 
 
 
 
 
 

 

answered Jan 31, 2013 by sreemathi.v
 
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