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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\sin^3x\;\cos^3x\]

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1 Answer

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Toolbox:
  • (i)$\sin^2x=1-\cos^2x.$
  • (ii)$\int x^ndx=\begin{bmatrix}x^{n+1}{n+1}\end{bmatrix}+c.$
  • (iii)Method of substitution:
  • If f(x)=t.
  • f'(x)dx=dt.
  • Thus $I=\int f(x)dx=\int t dt.$
Given $I=\int \sin^3x\cos ^3x dx=\int\sin^2x.\sin x.\cos^3xdx$.
 
Using the information in the tool box we can write as,
 
$I=\int \cos ^3x(1-\cos^2x)\sin xdx.$
 
Put $\cos x=t.$
 
On differentiating we get,
 
$-\sin xdx=dt$.
 
$\Rightarrow \sin xdx=-dt.$
 
On substituting t and dt,
 
Hence $I=\int t^3(1-t^2)(-dt).$
 
On multiplying we get,
 
$\;\;\;=-\int(t^3-t^5)dt.$
 
On integrating we get,
 
$\;\;\;=-\begin{bmatrix}\frac{t^4}{4}-\frac{t^6}{6}\end{bmatrix}+c.$
 
Substituting for t we get,
 
$\int\sin^3x\cos^3xdx=-\begin{bmatrix}\frac{\cos^4x}{4}-\frac{\cos^6x}{6}\end{bmatrix}+c.$

 

answered Jan 30, 2013 by sreemathi.v
 
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