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Home  >>  CBSE XII  >>  Math  >>  Integrals
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If $\large \frac{x^3dx}{\sqrt{1-x^2}}=\normalsize a(1-x^2)^{\Large\frac{3}{2}}+b\sqrt{1-x^2}+C$,then

\begin{array}{1 1}(A)\;a=\frac{1}{3},b=1 & (B)\;a=\frac{-1}{3},b=1\\(C)\;a=\frac{-1}{3},b=-1 &(D)\;a=\frac{1}{3},b=-1\end{array}

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1 Answer

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Toolbox:
  • If $f(x)$ is substituted by $f(t)$,then $f'(x)dx=f'(t)dt$
  • Hence $\int f(x)dx=\int f(t)dt.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c.$
Step 1:
Let $I=\int\large \frac{x^3dx}{\sqrt{1-x^2}}=\int\frac{x^2.xdx}{\sqrt{1-x^2}}$
Let $1-x^2=t^2$.On differentiating with respect to t we get,
$-2xdx=2tdt.$
$\Rightarrow xdx=-t dt$
If $1-x^2=t^2$,then $x^2=1-t^2$
On substituting this
$I=-\int\large\frac{(1-t^2)tdt}{t}$
$\;\;=-\int (1-t^2)dt.$
Step 2:
On integrating we get,
$-[t-\large\frac{t^3}{3}]$$+c$
Substituting $t=\sqrt{1-x^2}$ we get,
$-[\sqrt{1-x^2}-\large\frac{1}{3}$$(1-x^2)^{\Large\frac{3}{2}}]$$+c.$
$\Rightarrow \large\frac{1}{3}$$(1-x^2)^{\Large\frac{3}{2}}$$-\sqrt{1-x^2}+c.$
Therefore $a=\large\frac{1}{3}$,$b=-1$
Hence the correct answer is D.
answered May 23, 2013 by sreemathi.v
 

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