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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Evaluate $ \large \int \limits_{\normalsize 0}^{\frac{\pi}{2}}$$ \bigg (\sqrt (\tan x) + \large\frac{1}{(\sqrt \tan x) }$$\bigg)\;$$ dx$

(A) $\pi \quad$ (B) $\large\frac{2}{\sqrt \pi} \quad$ (C) $\pi \sqrt 2 \quad$ (D) $\large\frac{\sqrt \pi}{2}$

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1 Answer

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Substitute $\sin\;x - \cos\;x = t \rightarrow dt = (\cos\;x + \sin\;x)\; dx$
$\Rightarrow t^2 = \sin^2 x + \cos^2 x - 2 \sin\;x+\cos\;x = 1 - 2 \sin\;x \; \cos\;x$
$\Rightarrow \sin\;x\;\cos\;x = \large\frac{1-t^2}{2}$
Given $ \large \int_{\normalsize 0}^{\frac{\pi}{2}}$$ \bigg (\sqrt (\tan x) + \large\frac{1}{(\sqrt \tan x) }$$\bigg)\;$$ dx$
Since $\tan x = \large\frac{ \sin x}{ \cos x}$ $ \rightarrow \sqrt \tan x = \large \frac{\sqrt \sin x}{\sqrt \cos x}$
$\sqrt (\tan x )+ \large\frac{1}{(\sqrt \tan x)}$$ = \large \frac{\sqrt \sin x}{\sqrt \cos x}$ $ + \large \frac{\sqrt \cos x}{\sqrt \sin x}$$ = \large\frac{\sin x + \cos x }{\sqrt (\sin x \; \cos x)}$$ = \large\frac{dt}{dx}$$\times \large \frac{\sqrt 2}{\sqrt (1-t^2)}$
$\Rightarrow \large \int$$ \bigg (\sqrt (\tan x) + \large\frac{1}{(\sqrt \tan x) }$$\bigg)\; dx = \large\int $$ \large \frac{\sqrt 2 \; dt}{\sqrt (1 - t^2)}$
$ = \sqrt 2 \sin^{-1} t + c $$= \sqrt 2 \; \sin^{-1}(\sin x - \cos x) + c$
$ \Rightarrow \large \int_{\normalsize 0}^{\frac{\pi}{2}}$$ \bigg (\sqrt (\tan x) + \large\frac{1}{(\sqrt \tan x) }$$\bigg)\;$$ dx = \bigg [\sqrt 2 \; \sin^{-1}(\sin x - \cos x) \bigg]_{0}^{\frac{\pi}{2}}$
$\quad = \sqrt 2 \;[ \sin^{-1} 1 - \sin^{-1} (-1)] = \sqrt 2 \; \bigg ( \large\frac{\pi}{2} $$ + \large\frac{\pi}{2} $$\bigg)$
$\quad = \pi \sqrt 2$
answered Mar 20, 2014 by balaji.thirumalai
edited Mar 20, 2014 by balaji.thirumalai
 

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