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Evaluate the following $\int\frac{dx}{x\sqrt{x^4-1}}\;(Hint:Put\;x^2=\sec\theta)$

$\begin{array}{1 1} (A)\;\sec ^{-1} (x^2)+c \\ (B)\;\frac{1}{4} [\sec ^{-1} (x^2)]+c \\ (C)\;\frac{1}{6} [\sec ^{-1} (x^2)]+c \\ (D)\;\frac{1}{2} [\sec ^{-1} (x^2)]+c\end{array} $

1 Answer

  • $\int \sec^2 x dx=\tan x+c$
  • $\large\frac{d}{dx} $$(\sec x)=\sec x \tan x$
Let $I=\int\large\frac{dx}{x\sqrt{x^4-1}}$
Let $x^2=\sec \theta$
On differentiating with respect to x we get $2xdx=\sec \theta. \tan \theta.d \theta$
$dx=\large\frac{\sec \theta. \tan \theta.d \theta}{2x}$
On substituting for $x^2$ and dx we get,
$I=\large\int \frac{\sec \theta. \tan \theta . d\theta}{2 x^2 \sqrt {\sec ^2 \theta-1}}=\int \frac{\sec \theta. \tan \theta.d \theta}{2 \sec \theta \sqrt {\sec ^2 \theta-1}}$
But $\sec ^2 \theta-1=\tan ^2 \theta$
Therefore $I=\large\frac{1}{2} \int \frac{\tan \theta. d\theta}{\tan \theta}=\frac{1}{2} $$ \int d \theta$
On integrating we get $I=\frac{1}{2} \theta$
Since $ x^2=\sec \theta\; => \theta=\sec ^{-1} x^2$
Therefore $I=\frac{1}{2} [\sec ^{-1} (x^2)]+c$
answered Apr 16, 2013 by meena.p
edited Apr 11, 2016 by meena.p