∫0π/2 cot⁡xcot⁡x+tan⁡xdx=?

I=∫0π/2 cot⁡xcot⁡x+tan⁡xdx …..(i)=∫0π/2 cot⁡π2−xcot⁡π2−x+tan⁡π2−xdx=∫0π/2 tan⁡xtan⁡x+cot⁡xdx ……(ii)Now adding (i) and (ii), we get2I=∫0π/2 cot⁡x+tan⁡xtan⁡x+cot⁡xdx=[x]0π/2⇒ I=π4

∫0π/2 cot⁡xcot⁡x+tan⁡xdx=?

I=∫0π/2 cot⁡xcot⁡x+tan⁡xdx …..(i)=∫0π/2 cot⁡π2−xcot⁡π2−x+tan⁡π2−xdx=∫0π/2 tan⁡xtan⁡x+cot⁡xdx ……(ii)Now adding (i) and (ii), we get2I=∫0π/2 cot⁡x+tan⁡xtan⁡x+cot⁡xdx=[x]0π/2⇒ I=π4

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$\int_{0}^{π / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} d x = ?$

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$π$

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{3}$

answer is C.

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Detailed Solution

$I = \int_{0}^{π / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} d x$ …..(i)

$= \int_{0}^{π / 2} \frac{\sqrt{\cot (\frac{π}{2} - x)}}{\sqrt{\cot (\frac{π}{2} - x) + \sqrt{\tan (\frac{π}{2} - x)}}} d x$

$= \int_{0}^{π / 2} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} d x$ ……(ii)

Now adding (i) and (ii), we get

$\begin{array}{l} 2 I = \int_{0}^{π / 2} \frac{\sqrt{\cot x} + \sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} d x = [x]_{0}^{π / 2} \\ \Rightarrow I = \frac{π}{4} \end{array}$