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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $(x^3-1)^\frac{1}{3}\;x^5$

$\begin{array}{1 1}\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c \\ \frac{1}{5}(x^3-1)^\frac{5}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c \\ \frac{1}{7}(x^2-1)^\frac{7}{3}+\frac{1}{4}(x^2-1)^\frac{4}{3}+c\\ \frac{1}{5}(x^2-1)^\frac{5}{3}+\frac{1}{4}(x^2-1)^\frac{4}{3}+c\end{array}$

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Toolbox:
  • 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 $\frac{dx}{dt}=g'(t).$
  • dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int(x^3-1).x^5$.
 
This can be written as
 
$\;\;\;=\int (x^3-1)\frac{1}{3}.x^3.x^2dx$.
 
Let $x^3-1=t\Rightarrow x^3=(t+1).$
 
$3x^2dx=dt \Rightarrow x^2dx=\frac{dt}{3}$.
 
Now substituting for x and dx we get,
 
$I=\int (t)^\frac{1}{3}.(t+1).\frac{dt}{3}$.
 
\;\;\;=\frac{1}{3}\int(t^\frac{4}{3}+t^\frac{1}{3})dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{3}\begin{bmatrix}\frac{t^\frac{7}{3}}{\frac{7}{3}}+\frac{t^\frac{4}{3}}{\frac{4}{3}}\end{bmatrix}+c.$
 
$\;\;\;=\frac{1}{3}[\frac{3}{7}(t^\frac{7}{3})+\frac{3}{4}(t^\frac{4}{3})]+c.$
 
Substituting back for t we get,
 
$\;\;\;=\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c.$
 
Hence $\int(x^3-1).x^5=\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c.$
 
 
 

 

answered Jan 28, 2013 by sreemathi.v
 
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