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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $ \int \sqrt {\tan x }+ \sqrt {\cot x }dx$

$(a)\;\frac{1}{\sqrt 2} \tan^{-1} \bigg(\frac{\sqrt {\tan x}}{2}-\frac{1}{\sqrt {2 \tan x }}\bigg)+c \qquad(b)\;\tan ^{-1} (\sqrt {\tan x}+\sqrt {2 \tan x }) \qquad(c)\;x+\sqrt {2x}+c \qquad (d)\;None$

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1 Answer

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Suppose $\tan x =t => \tan x =t^2 $
differentiate with x :
$\sec ^2 x d= 2t dt$
$ \int \large\frac{1+t^2}{t} \times \frac{2t.dt}{(1+t^4)}$
$\qquad=2 \int \large\frac{t^2(1+1/t^2)}{t^2 (t^2+1/t^2)} .dt$
$\qquad=2 \int \large\frac{(1+1/t^2)}{(t^2+1/t^2)} .dt$
$\qquad=2 \int \large\frac{1+1/t^2}{(t-\frac{1}{t})^2+(\sqrt 2 )^2}$$ .dt$
$t-\large\frac{1}{t}$$ =A$ => differentiate with t.
$1+\large\frac{1}{t^2}$$dt=dA$
$\qquad= 2 \int \large\frac{dA}{(A)^2 +{\sqrt 2)^2}}$
$\qquad =\large\frac{1}{\sqrt 2} \tan ^{-1} \bigg( \large\frac{A}{\sqrt 2}\bigg)+c$
Put value of A and after put value of
$\large\frac{1}{\sqrt 2} $$\tan^{-1} \bigg(\frac{\sqrt {\tan x}}{2}-\large\frac{1}{\sqrt {2 \tan x }}\bigg)$$+c$
Hence a is the correct answer.
answered Dec 23, 2013 by meena.p
 
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