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# Evaluate $\large \int \limits_{\normalsize 0}^{\frac{\pi}{2}}$$\bigg (\sqrt (\tan x) + \large\frac{1}{(\sqrt \tan x) }$$\bigg)\;$$dx (A) \pi \quad (B) \large\frac{2}{\sqrt \pi} \quad (C) \pi \sqrt 2 \quad (D) \large\frac{\sqrt \pi}{2} Can you answer this question? ## 1 Answer 0 votes Substitute \sin\;x - \cos\;x = t \rightarrow dt = (\cos\;x + \sin\;x)\; dx \Rightarrow t^2 = \sin^2 x + \cos^2 x - 2 \sin\;x+\cos\;x = 1 - 2 \sin\;x \; \cos\;x \Rightarrow \sin\;x\;\cos\;x = \large\frac{1-t^2}{2} Given \large \int_{\normalsize 0}^{\frac{\pi}{2}}$$ \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$