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Question Number 83146 by 09658867628 last updated on 28/Feb/20

∫_( 0) ^(π/2)   ((√(cot x))/((√(cot x)) + (√(tan x)))) dx =

π/20cotxcotx+tanxdx=

Answered by Kunal12588 last updated on 28/Feb/20

I=∫_0 ^( π/2) ((√(cot x))/((√(cot x))+(√(tan x))))dx  ⇒I=∫_0 ^( π/2) ((√(tan x))/((√(tan x))+(√(cot x))))dx  ⇒I=(1/2)∫_0 ^( π/2) dx  ⇒I=(1/2)×(π/2)=(π/4)  ∫_( 0) ^(π/2)   ((√(cot x))/((√(cot x)) + (√(tan x)))) dx =(π/4)

I=0π/2cotxcotx+tanxdxI=0π/2tanxtanx+cotxdxI=120π/2dxI=12×π2=π4π/20cotxcotx+tanxdx=π4

Answered by niroj last updated on 28/Feb/20

 let, I= ∫_0 ^(π/2)  (( (√(cot x)))/( (√(cot x)) +(√(tan x))))dx.....(i)      = ∫_0 ^(π/2)   ((√( cot ((π/2)−x)))/( (√(cot ((π/2))) −x)+ (√(tan ((π/2) −x)))))dx     ∵ ∫_0 ^( a) xdx=∫_0 ^a (a−x)dx   = ∫_0 ^( (π/2))  ((√(tan x))/( (√(tan x)) +(√(cot x))))dx......(ii)    added (i)&(ii)    2I= ∫_0 ^(π/2) (  ((   (√(cot x)))/( (√(cot x))  +(√(tan x)))) + ((√(tan x))/(  (√(tan x)) +(√(cot x)))))dx    2I= ∫_0 ^( (π/2))   (((  (√(cot x))  +(√(tan x)))/(  (√(cot x))+(√(tan x)))))dx    2I= ∫_0 ^(π/2)  dx    2I= [ x]_0 ^(π/2)     2I= ((π/2)−0)      2I= (π/2) ⇒ I= (π/4) //.

let,I=0π2cotxcotx+tanxdx.....(i)=0π2cot(π2x)cot(π2x)+tan(π2x)dx0axdx=0a(ax)dx=0π2tanxtanx+cotxdx......(ii)added(i)&(ii)2I=0π2(cotxcotx+tanx+tanxtanx+cotx)dx2I=0π2(cotx+tanxcotx+tanx)dx2I=0π2dx2I=[x]0π22I=(π20)2I=π2I=π4//.

Commented by peter frank last updated on 28/Feb/20

thank you both

thankyouboth

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