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tan-1-1-sinx-1-sinx-dx-




Question Number 45669 by arvinddayama01@gmail.com last updated on 15/Oct/18
∫tan^(−1) (√((1−sinx)/(1+sinx))) dx=?
tan11sinx1+sinxdx=?
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
t=tan(x/2)       now tan^(−1) (√((1−((2t)/(1+t^2 )))/(1+((2t)/(1+t^2 )))))   =(√(((1−t)^2 )/((1+t)^2 ))) =((1−t)/(1+t))=((1−tan(x/2))/(1+tan(x/2)))=tan((π/4)−(x/2))  so tan^(−1) {tan((π/4)−(x/2))}=(π/4)−(x/2)  ∫(π/4)−(x/2)  dx  =((πx)/4)−(x^2 /4)+c
t=tanx2nowtan112t1+t21+2t1+t2=(1t)2(1+t)2=1t1+t=1tanx21+tanx2=tan(π4x2)sotan1{tan(π4x2)}=π4x2π4x2dx=πx4x24+c
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
most welcome...
mostwelcome
Commented by arvinddayama01@gmail.com last updated on 15/Oct/18
Thanku very much
Thankuverymuch
Answered by vermasir last updated on 16/Oct/18
∫tan^(−1) (√((1−cos((π/2)−x))/(1+cos((π/2)+x))))  ∫tan^(−1) (√((2sin^2 ((π/4)−(x/2)))/(2cos^2 ((π/4)−(x/2)))))  ∫tan^(−1) ∣tan((π/4)−(x/2))∣dx  ∫((π/4)−(x/2))dx  (π/4)x−(x^2 /4)+c
tan11cos(π2x)1+cos(π2+x)tan12sin2(π4x2)2cos2(π4x2)tan1tan(π4x2)dx(π4x2)dxπ4xx24+c

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