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

Integrate : $\int \large\frac{\sin x +\cos x }{\sqrt {\sin 2x}}$$dx$

$(a)\;\frac{-1}{2}\tan ^{-1} (\sqrt {\frac{\tan x }{2}})+c \\(b)\;\frac{1}{2}\tan ^{-1} (\tan x)+c \\(c)\;\frac{1}{2}\tan ^{-1} (\sqrt {\frac{-1}{\sqrt {2 \tan x }}})+c \\ (d)\;\frac{1}{2}\tan ^{-1} (\sqrt {\frac{\tan x }{2}}-\frac{1}{\sqrt {2 \tan x }})+c$

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1 Answer

$\int \large\frac{\sin x +\cos x }{\sqrt {2 \sin x \cos x }}$$dx$
$\qquad= \int \large\frac{\sqrt {\sin x }}{\sqrt {2 \cos x }}+\frac{\sqrt {\cos x }}{\sqrt {2 \sin x}}$$dx$
$\qquad= \large\frac{1}{\sqrt 2}$$ \int (\sqrt {\tan x}+\sqrt {\cot x} )dx$
$\large\frac{1}{2}$$\tan ^{-1} (\sqrt {\large \frac{\tan x }{2}}-\frac{1}{\sqrt {2 \tan x }})+c$
Hence d is the correct answer.
answered Dec 30, 2013 by meena.p
 

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