Question Number 113910 by bobhans last updated on 16/Sep/20

(1)∫_0 ^π  ((sin ^4 x)/((1+cos x)^2 )) dx ?  (2) lim_(x→∞)  ((√(1−cos (((2π)/x))))/(1/x)) ?

Answered by bemath last updated on 16/Sep/20

(2) set (1/x) = z , z→0   lim_(z→0)  ((√(1−cos 2πz))/z) = lim_(z→0)  ((√(1−(1−2sin ^2 πz)))/z)  = lim_(z→0)  (((√2) sin πz)/z) = π(√2)

Answered by 1549442205PVT last updated on 16/Sep/20

(1)I=∫_0 ^π  ((sin ^4 x)/((1+cos x)^2 )) dx   Since 1+cosx=2cos^2 (x/2) and sinx=2sinxcosx  ((sin^4 x)/((1+cosx)^4 ))=((16sin^4 (x/2)cos^4 (x/2))/(4cos^4 (x/2)))=4sin^4 (x/2)  Hence,I=4∫_0 ^( π) sin^4 (x/2)dx=2∫_0 ^( π) (1−cosx)^2 dx  =2∫_0 ^( π) (1−2cosx+cos^2 x)dx  =2∫_0 ^( π) dx−4∫_0 ^( π) cosx+∫_0 ^( π) (1+cos2x)dx  =(3x−4sinx+(1/2)sin2x)∣_0 ^π   =3π

Answered by MJS_new last updated on 16/Sep/20

∫((sin^4  x)/((1+cos x)^2 ))dx=∫(((1−cos^2  x)^2 )/((1+cos x)^2 ))dx=  =∫(((1−cos x)^2 (1+cos x)^2 )/((1+cos x)^2 ))dx=  =∫(1−2cos x +cos^2  x)dx=  =x−2sin x+(x/2)+((cos x sin x)/2)=  =(3/2)x−(1/2)(4−cos x)sin x +C  ⇒ ∫_0 ^π ((sin^4  x)/((1+cos x)^2 ))dx=((3π)/2)

Answered by mathmax by abdo last updated on 17/Sep/20

1) A =∫_0 ^π  ((sin^4 x)/((1+cosx)^2 ))dx ⇒A =∫_0 ^π  (((2sin((x/2))cos((x/2)))^4 )/((2cos^2 ((x/2)))^2 )) dx  =4∫_0 ^π   ((sin^4 ((x/2)) cos^4 ((x/2)))/(cos^4 ((x/2)))) dx =4 ∫_0 ^π  sin^4 ((x/2))dx  =_((x/2)=t)    8∫_0 ^(π/2)  sin^4 (t) dt =8 ∫_0 ^(π/2) (((1−cos(2t))/2))^2  dt  =2 ∫_0 ^(π/2) (1−2cos(2t) +cos^2 (2t))dt  =2 [t−sin(2t)]_0 ^(π/2)  +∫_0 ^(π/2) (1+cos(4t))dt  =2{(π/2)} +(π/2) =π +(π/2) ⇒A =((3π)/2)