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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Choose the correct answer in $\int\limits_0^\frac{2}{3}\frac{dx}{4+9x^2}$ equals

$\begin{array}{1 1}(A)\;\frac{\pi}{6} \\ (B)\;\frac{\pi}{12} \\ (C)\frac{\pi}{24} \\ (D)\frac{\pi}{4} \end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
  • (ii)$\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1} (x/a)$
$\int\limits_0^\frac{2}{3}\large\frac{dx}{4+9x^2}\;$
 
This can be written as
 
$\int\limits_0^\frac{2}{3}\large\frac{dx}{9(x^2+\frac{4}{9})}\;$
 
$=\frac{1}{9}\int\limits_0^\frac{2}{3}\large\frac{dx}{x^2+(2/3)^2}\;$
 
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1} (x/a)$
 
$=\frac{1}{9}\int\limits_0^\frac{2}{3}\large\frac{dx}{x^2+(2/3)^2}\;$
 
$=\bigg[\frac{1}{9} \times \frac{1}{2/3} \tan^{-1}(\frac{x}{2/3})\bigg]_0^{\frac{2}{3}}$
 
On applying limits,
 
$=\bigg[\frac{1}{9} \times \frac{3}{2} \tan^{-1}(\frac{3x}{2})\bigg]_0^{\frac{2}{3}}$
 
$=\large\frac{1}{6}\bigg[\tan ^{-1}\bigg(\frac{3}{2} \times \frac {2}{3} \bigg)-\tan ^{-1}(0)\bigg]$
 
$=\large \frac{1}{6} \tan ^{-1}(1)-0$
 
But $\tan^{-1}(1)=\frac{\pi}{4}\;Therefore\; \pi=\frac{1}{6} \times \frac{\pi}{4}=\frac{\pi}{24}$
 
Hence the correct answer is C

 

answered Feb 12, 2013 by meena.p
 
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