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Question Number 162471 by mr W last updated on 29/Dec/21

[reposted]  find ∫_( 0) ^( (𝛑/2))  sin^8 (x)dx + ∫_( 0) ^( 1)  sin^(-1) ((x)^(1/8) ) dx=?

[reposted]findβˆ«Ο€20sin8(x)dx+∫10sinβˆ’1(x8)dx=?

Answered by Ar Brandon last updated on 30/Dec/21

I=∫_0 ^(Ο€/2) sin^8 xdx+∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) )dx  I_1 =∫_0 ^(Ο€/2) sin^8 xdx , I_2 =∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) )dx  I_2 =∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) )dx , x=t^8 β‡’dx=8t^7 dt       =8∫_0 ^1 t^7 sin^(βˆ’1) (t)dt,  { ((u(t)=sin^(βˆ’1) (t))),((vβ€²(t)=t^7 )) :}β‡’ { ((uβ€²(t)=(1/( (√(1βˆ’t^2 )))))),((v(t)=(1/8)t^8 )) :}       =[t^8 sin^(βˆ’1) (t)]_0 ^1 βˆ’βˆ«_0 ^1 (t^8 /( (√(1βˆ’t^2 ))))dt, t=sinΟ‘β‡’dt=cosΟ‘dΟ‘       =(Ο€/2)βˆ’βˆ«^(Ο€/2) _0 sin^8 Ο‘dΟ‘=(Ο€/2)βˆ’βˆ«_0 ^(Ο€/2) sin^8 xdx=(Ο€/2)βˆ’I_1   I=I_1 +I_2 =I_1 +(Ο€/2)βˆ’I_1 =(Ο€/2)

I=∫0Ο€2sin8xdx+∫01sinβˆ’1(x8)dxI1=∫0Ο€2sin8xdx,I2=∫01sinβˆ’1(x8)dxI2=∫01sinβˆ’1(x8)dx,x=t8β‡’dx=8t7dt=8∫01t7sinβˆ’1(t)dt,{u(t)=sinβˆ’1(t)vβ€²(t)=t7β‡’{uβ€²(t)=11βˆ’t2v(t)=18t8=[t8sinβˆ’1(t)]01βˆ’βˆ«01t81βˆ’t2dt,t=sinΟ‘β‡’dt=cosΟ‘dΟ‘=Ο€2βˆ’βˆ«0Ο€2sin8Ο‘dΟ‘=Ο€2βˆ’βˆ«0Ο€2sin8xdx=Ο€2βˆ’I1I=I1+I2=I1+Ο€2βˆ’I1=Ο€2

Commented by Ar Brandon last updated on 30/Dec/21

Thanks Sir. Corrected ! Itβ€²s okay now.

ThanksSir.Corrected!Itβ€²sokaynow.

Commented by Ar Brandon last updated on 29/Dec/21

Commented by Ar Brandon last updated on 29/Dec/21

No, Sir.

No,Sir.

Commented by mr W last updated on 30/Dec/21

thanks for confirming sir!

thanksforconfirmingsir!

Answered by mr W last updated on 29/Dec/21

we can solve this integral without  calculating it.

wecansolvethisintegralwithoutcalculatingit.

Commented by mr W last updated on 29/Dec/21

we want to find I=I_1 +I_2  with   I_1 =∫_0 ^(Ο€/2) sin^8  x dx   I_2 =∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) ) dx  now we apply method above to find  I_2 =∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) )dx  y=f(x)=sin^(βˆ’1) ((x)^(1/8) )   β‡’x=sin^8  (y)=f^(βˆ’1) (y)  x∈[0,1] β‡’y∈[0,(Ο€/2)]  I_2 =∫_0 ^1 sin^(βˆ’1) ((x)^(1/8) )dx       =1Γ—(Ο€/2)βˆ’βˆ«_0 ^(Ο€/2) sin^8  (y)dy       =(Ο€/2)βˆ’βˆ«_0 ^(Ο€/2) sin^8  (x)dx       =(Ο€/2)βˆ’I_1   I=I_1 +I_2 =I_1 +(Ο€/2)βˆ’I_1 =(Ο€/2) βœ“  this works even when we donβ€²t know  how to calculate the concrete integrals.    for the same reason we can get e.g.  ∫_0 ^(Ο€/2) sin^(100)  x dx+∫_0 ^1 sin^(βˆ’1) ((x)^(1/(100)) ) dx=(Ο€/2)

wewanttofindI=I1+I2withI1=∫0Ο€2sin8xdxI2=∫01sinβˆ’1(x8)dxnowweapplymethodabovetofindI2=∫01sinβˆ’1(x8)dxy=f(x)=sinβˆ’1(x8)β‡’x=sin8(y)=fβˆ’1(y)x∈[0,1]β‡’y∈[0,Ο€2]I2=∫01sinβˆ’1(x8)dx=1Γ—Ο€2βˆ’βˆ«0Ο€2sin8(y)dy=Ο€2βˆ’βˆ«0Ο€2sin8(x)dx=Ο€2βˆ’I1I=I1+I2=I1+Ο€2βˆ’I1=Ο€2βœ“thisworksevenwhenwedonβ€²tknowhowtocalculatetheconcreteintegrals.forthesamereasonwecangete.g.∫0Ο€2sin100xdx+∫01sinβˆ’1(x100)dx=Ο€2

Commented by mr W last updated on 29/Dec/21

Commented by mr W last updated on 29/Dec/21

geometrically ∫_0 ^a f(x)dx means the  area under the curve y=f(x). i.e.  ∫_0 ^a ydx=∫_0 ^a f(x)dx=A_1   we see A_1 =abβˆ’A_2  while A_2  is the  area under the curve x=f^(βˆ’1) (y), i.e.  A_2 =∫_0 ^b xdy=∫_0 ^b f^(βˆ’1) (y)dy.  therefore we can get  ∫_0 ^a f(x)dx=abβˆ’βˆ«_0 ^b f^(βˆ’1) (y)dy

geometrically∫0af(x)dxmeanstheareaunderthecurvey=f(x).i.e.∫0aydx=∫0af(x)dx=A1weseeA1=abβˆ’A2whileA2istheareaunderthecurvex=fβˆ’1(y),i.e.A2=∫0bxdy=∫0bfβˆ’1(y)dy.thereforewecanget∫0af(x)dx=abβˆ’βˆ«0bfβˆ’1(y)dy

Commented by Tawa11 last updated on 30/Dec/21

Wow great sir.

Wowgreatsir.

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