Question Number 115030 by bobhans last updated on 23/Sep/20

∫_(−(π/2)) ^(π/2) (√(sec x−cos x)) dx =?

Answered by bemath last updated on 23/Sep/20

∫_(−(π/2)) ^(π/2)  (√((1−cos ^2 x)/(cos x))) dx = ∫_(−(π/2)) ^(π/2)  ((∣sin x∣)/( (√(cos x)))) dx  = −∫_(−(π/2)) ^0 ((sin x)/( (√(cos x)))) dx + ∫_0 ^(π/2)  ((sin x)/( (√(cos x)))) dx   = ∫_(−(π/2)) ^0 ((d(cos x))/( (√(cos x)))) −∫_0 ^(π/2)  ((d(cos x))/( (√(cos x))))  = 2 [ (√(cos x )) ]_(−(π/2)) ^(   0)   −2 [ (√(cos x )) ] _0^(π/2)   = 2 −2(0−1) = 4

Answered by mathmax by abdo last updated on 23/Sep/20

A =∫_(−(π/2)) ^(π/2) (√((1/(cosx))−cosx))dx  ⇒A =2∫_0 ^(π/2) (√((1−cos^2 x)/(cosx)))dx  =2 ∫_0 ^(π/2) ((∣sinx∣)/(√(cosx)))dx =2 ∫_0 ^(π/2)  ((sinx)/(√(cosx)))dx =2[−2(√(cosx))]_0 ^(π/2)  =2{2} =4