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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\sqrt{\sin2x}\cos2x\]

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Toolbox:
  • (i)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 $I=\int f(g(t).g'(t))dt.$
  • (ii)$\int \cos axdx=\frac{1}{a}\sin a x.$
Given $I=\int \sqrt{\sin 2x}.\cos 2xdx.$
 
Put $\sin 2x=t.$
 
$2\cos2xdx=dt.\Rightarrow\cos 2xdx=\frac{dt}{2}.$
 
Now substituting for x and dx we get,
 
$I=2\int\sqrt t \frac{dt}{2}=\frac{1}{2}\int\sqrt t dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}\begin{bmatrix}\frac{t^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}+c.$
 
$\;\;\;=\frac{1}{3}\big(t^\frac{3}{2}\big)+c.$
 
Substituting back for t we get,
 
$\int\sqrt{sin 2x}\cos 2x dx=\frac{1}{3}(\sin2x)^\frac{3}{2}+c.$
 
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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