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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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integrate : $\int \sec^3 x dx$

\[\begin {array} {1 1} (a)\;\frac{1}{2}\{ \tan x \sec x - \log (\tan x - \sec x)\}+c \\ (b)\;\frac{1}{2}\{ \tan x \sec x + \log (\tan x + \sec x)\}+c \\ (c)\;\frac{1}{2} \{ \tan x \sec x + \log (\tan x - \sec x)\}+c \\ (d)\;None \end {array}\]
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$ \int \sec x . \sec^2 x dx$
=> $ \int \sqrt{1+\tan ^2 x}. \sec^2. x dx$
$\tan x =t=> \sec^2 x dx=dt$
=> $\int \sqrt {1+t^2}.dt$
=> $\large\frac{1}{2}$$ t \sqrt {1+t^2} + \large\frac{1}{2}$$ \log \{ t+ \sqrt {1+t^2}\}+c$
=> $\large\frac{1}{2}$$ \tan x . \sec x +\large\frac{1}{2} $$ \log (\tan x + \sec x \}+c$
Hence b is the correct answer.
answered Dec 19, 2013 by meena.p
 
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