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Question Number 101272 by bemath last updated on 01/Jul/20

∫(√(sec x)) dx

secxdx

Commented by john santu last updated on 01/Jul/20

I= ∫(dx/(√(cos x))) = ∫ ((1+sin x−sin x)/(√(cos x))) dx  I_1  = ∫ (((cos (x/2)−sin (x/2))^2  dx)/(√(cos^2 ((x/2))−sin^2 ((x/2)))))  I_1  = −2(cos (x/2)+sin (x/2))^(3/2)   I_2  = −∫ ((sin x)/(√(cos x))) dx = ∫ ((d(cos x))/(√(cos x)))  = 2(√(cos x))  I = I_1  + I_2  = 2(√(cos x))−2(cos (x/2)+sin (x/2))^(3/2) + c

I=dxcosx=1+sinxsinxcosxdxI1=(cosx2sinx2)2dxcos2(x2)sin2(x2)I1=2(cosx2+sinx2)32I2=sinxcosxdx=d(cosx)cosx=2cosxI=I1+I2=2cosx2(cosx2+sinx2)32+c

Commented by MJS last updated on 01/Jul/20

sorry but I_1 ≠−2(cos (x/2) +sin (x/2))^(3/2)   because (((c−s)^2 )/(√(c^2 −s^2 )))=(((c−s)^(3/2) )/(√(c+s))) and I don′t see  how to get your solution

sorrybutI12(cosx2+sinx2)3/2because(cs)2c2s2=(cs)3/2c+sandIdontseehowtogetyoursolution

Answered by mathmax by abdo last updated on 01/Jul/20

this integral is not solvable (elliptic)

thisintegralisnotsolvable(elliptic)

Commented by MJS last updated on 01/Jul/20

yes you are right

yesyouareright

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