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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \cos ^{-1} \bigg(\large\frac{1-\tan^2 x}{1+\tan^2x}\bigg) $$dx$

$(a)\;x+c \\(b)\;x^2+c \\(c)\; x^3+c \\ (d)\;x^4+c $

1 Answer

We know : $ \large\frac{1-\tan ^2 x}{1+\tan ^2 x}$$=\cos 2x$
$\int \cos^{-1} (\cos 2x).dx$
$\qquad=\int 2x.dx$
$\qquad= x^2 +c$
Hence b is the correct answer.
answered Dec 31, 2013 by meena.p
 
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