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Question Number 54971 by peter frank last updated on 15/Feb/19

∫_0 ^(π/2) (((√(tanx)) sin^2 x)/((√(sin xcos x)) tan x))dx

0π2tanxsin2xsinxcosxtanxdx

Answered by kaivan.ahmadi last updated on 15/Feb/19

∫_0 ^(π/2) (((√((sinx)/(cosx)))sin^2 x)/((√(sinxcosx))((sinx)/(cosx))))dx=  ∫_0 ^(π/2) sinxdx=−cosx]_0 ^(π/2) =1

0π2sinxcosxsin2xsinxcosxsinxcosxdx=0π2sinxdx=cosx]0π2=1

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19

∫(((√(tanx)) ×tan^2 x×cos^2 x)/((√(tanx×cos^2 x)) ×tanx))dx  ∫(((√(tanx)) ×tanx×cosx)/((√(tanx)) ))dx  ∫sinxdx  =−∣cosx∣_0 ^(π/2)    =−(0−1)=1ans

tanx×tan2x×cos2xtanx×cos2x×tanxdxtanx×tanx×cosxtanxdxsinxdx=cosx0π2=(01)=1ans

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