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Question Number 86313 by john santu last updated on 28/Mar/20

∫_0 ^(π/2)  (dx/(√(1+tan x)))

π20dx1+tanx

Commented by john santu last updated on 28/Mar/20

((√(sin x))/(√(sin x+cos x))) + ((√(cos x))/(√(sin x+cos x))) = 1  how ?

sinxsinx+cosx+cosxsinx+cosx=1how?

Commented by john santu last updated on 28/Mar/20

(√(cos x)) + (√(sin x)) = (√(sin x+cos x)) ?

cosx+sinx=sinx+cosx?

Answered by TANMAY PANACEA. last updated on 28/Mar/20

i think...  ∫_0 ^(π/2) (dx/(1+((√(sinx))/(√(cosx)))))=∫_0 ^(π/2) ((√(cosx))/((√(sinx)) +(√(cosx))))dx  using ∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx  I=∫_0 ^(π/2) ((√(cos((π/2)−x)))/((√(sin((π/2)−x))) +(√(cos((π/2)−x)))))dx  =∫_0 ^(π/2) ((√(sinx))/((√(cosx)) +(√(sinx))))dx  2I=∫_0 ^(π/2) ((√(cosx))/((√(sinx)) +(√(cosx))))+((√(sinx))/((√(cosx)) +(√(sinx))))dx  2I=∫_0 ^(π/2) dx  I=(π/2)×(1/2)=(π/4)

ithink...0π2dx1+sinxcosx=0π2cosxsinx+cosxdxusing0af(x)dx=0af(ax)dxI=0π2cos(π2x)sin(π2x)+cos(π2x)dx=0π2sinxcosx+sinxdx2I=0π2cosxsinx+cosx+sinxcosx+sinxdx2I=0π2dxI=π2×12=π4

Commented by jagoll last updated on 28/Mar/20

king formula

kingformula

Commented by TANMAY PANACEA. last updated on 28/Mar/20

thank you sir

thankyousir

Commented by john santu last updated on 28/Mar/20

it does mean the equation wrong sir?

itdoesmeantheequationwrongsir?

Commented by TawaTawa1 last updated on 28/Mar/20

Sir tanmay, please help with  Q86324.

Sirtanmay,pleasehelpwithQ86324.

Answered by MJS last updated on 28/Mar/20

∫(dx/(√(1+tan x)))=       [t=(√(1+tan x)) → dx=2cos^2  x (√(1+tan x))dt]  =2∫(dt/(t^4 −2t^2 +2))  and now it′s easy

dx1+tanx=[t=1+tanxdx=2cos2x1+tanxdt]=2dtt42t2+2andnowitseasy

Commented by jagoll last updated on 28/Mar/20

= 2 ∫ (dt/((t^2 −1)^2 +1))   let t^2 −1 = tan w⇒ 2t dt = sec^2 w dw  = ∫ ((sec^2 w dw)/(sec^2 w)) = w + c  = tan^(−1) (t^2 −1) + c  = tan^(−1) (tan (x)) + c  it correct sir?

=2dt(t21)2+1lett21=tanw2tdt=sec2wdw=sec2wdwsec2w=w+c=tan1(t21)+c=tan1(tan(x))+citcorrectsir?

Commented by MJS last updated on 28/Mar/20

no  w=arctan (t^2 −1) → dt=((t^4 −2t^3 +2)/(2t))dw  ⇒ ∫(dw/(√(1+tan w))) so you′re back at the start

now=arctan(t21)dt=t42t3+22tdwdw1+tanwsoyourebackatthestart

Commented by MJS last updated on 28/Mar/20

we must decompose  2∫(dt/(t^4 −2t^2 +2))=  =∫((at+b)/(t^2 −(√(2+2(√2)))t+(√2)))dt+∫((ct+d)/(t^2 +(√(2+2(√2)t+(√2)))))dt=  =−((√(−1+(√2)))/2)∫((t−(√(2+2(√2))))/(t^2 −(√(2+2(√2)))t+(√2)))dt+((√(−1+(√2)))/2)∫((t+(√(2+2(√2))))/(t^2 +(√(2+(√2)))t+(√2)))dt  now use formula

wemustdecompose2dtt42t2+2==at+bt22+22t+2dt+ct+dt2+2+22t+2dt==1+22t2+22t22+22t+2dt+1+22t+2+22t2+2+2t+2dtnowuseformula

Commented by jagoll last updated on 28/Mar/20

waw...yes sir. i try solve it

waw...yessir.itrysolveit

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