Question

integral of 0 to pi/2 root (cot x) / root( cotx)+root (tan x) dx​

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \sf\int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

FORMULA TO BE IMPLEMENTED

$\boxed{ \: \: \displaystyle \sf\int\limits_{0}^{ a } f(x) \, dx = \int\limits_{0}^{ a } f(a - x) \, dx \: }$

EVALUATION

Let I be the given given integral

$\displaystyle \sf I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx \: \: \: - - -( 1)$

We now use the formula

$\boxed{ \: \: \displaystyle \sf\int\limits_{0}^{ a } f(x) \, dx = \int\limits_{0}^{ a } f(a - x) \, dx \: }$

$\displaystyle \sf I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot \bigg( \dfrac{\pi}{2} - x \bigg)} }{ \sqrt{ \cot \bigg( \dfrac{\pi}{2} - x \bigg)} + \sqrt{ \tan \bigg( \dfrac{\pi}{2} - x \bigg)} } \, dx$

$\displaystyle \sf \implies \: I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \tan x} }{ \sqrt{ \tan x} + \sqrt{ \cot x} } \, dx \: \: \:$

$\displaystyle \sf \implies \: I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx \: \: \: - - -( 2)$

Now Equation 1 + Equation 2 gives

$\displaystyle \sf 2I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx + \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \tan x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

$\displaystyle \sf \implies \: 2I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} + \sqrt{ \tan x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

$\displaystyle \sf \implies \: 2I = \int\limits_{0}^{ \frac{\pi}{2} } 1 \, dx$

$\displaystyle \sf \implies \: 2I =x \bigg|_{0}^{ \frac{\pi}{2} }$

$\displaystyle \sf \implies \: 2I = \frac{\pi}{2} - 0$

$\displaystyle \sf \implies \: 2I = \frac{\pi}{2}$

$\displaystyle \sf \implies \: I = \frac{\pi}{4}$

FINAL ANSWER

$\displaystyle \sf \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx = \frac{\pi}{4}$

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Answer 2

Step-by-step explanation:

SOLUTION

TO EVALUATE

$\displaystyle \sf\int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

FORMULA TO BE IMPLEMENTED

$\boxed{ \: \: \displaystyle \sf\int\limits_{0}^{ a } f(x) \, dx = \int\limits_{0}^{ a } f(a - x) \, dx \: }$

EVALUATION

Let I be the given given integral

$\displaystyle \sf I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx \: \: \: - - -( 1)$

We now use the formula

$\boxed{ \: \: \displaystyle \sf\int\limits_{0}^{ a } f(x) \, dx = \int\limits_{0}^{ a } f(a - x) \, dx \: }$

$\displaystyle \sf I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot \bigg( \dfrac{\pi}{2} - x \bigg)} }{ \sqrt{ \cot \bigg( \dfrac{\pi}{2} - x \bigg)} + \sqrt{ \tan \bigg( \dfrac{\pi}{2} - x \bigg)} } \, dx$

$\displaystyle \sf \implies \: I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \tan x} }{ \sqrt{ \tan x} + \sqrt{ \cot x} } \, dx \: \: \:$

$\displaystyle \sf \implies \: I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx \: \: \: - - -( 2)$

Now Equation 1 + Equation 2 gives

$\displaystyle \sf 2I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx + \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \tan x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

$\displaystyle \sf \implies \: 2I = \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} + \sqrt{ \tan x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx$

$\displaystyle \sf \implies \: 2I = \int\limits_{0}^{ \frac{\pi}{2} } 1 \, dx$

$\displaystyle \sf \implies \: 2I =x \bigg|_{0}^{ \frac{\pi}{2} }$

$\displaystyle \sf \implies \: 2I = \frac{\pi}{2} - 0$

$\displaystyle \sf \implies \: 2I = \frac{\pi}{2}$

$\displaystyle \sf \implies \: I = \frac{\pi}{4}$

FINAL ANSWER

$\displaystyle \sf \int\limits_{0}^{ \frac{\pi}{2} } \frac{ \sqrt{ \cot x} }{ \sqrt{ \cot x} + \sqrt{ \tan x} } \, dx = \frac{\pi}{4}$

━━━━━━━━━━━━━━━━

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