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reposted-find-0-2-sin-8-x-dx-0-1-sin-1-x-1-8-dx-




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=?
$$\left[{reposted}\right] \\ $$$${find}\:\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}}\:\mathrm{sin}^{\mathrm{8}} \left(\mathrm{x}\right){dx}\:+\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{sin}^{-\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{\mathrm{x}}\right)\:{dx}=? \\ $$
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)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} {xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right){dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} {xdx}\:,\:{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right){dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right){dx}\:,\:{x}={t}^{\mathrm{8}} \Rightarrow{dx}=\mathrm{8}{t}^{\mathrm{7}} {dt} \\ $$$$\:\:\:\:\:=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{7}} \mathrm{sin}^{βˆ’\mathrm{1}} \left({t}\right){dt},\:\begin{cases}{{u}\left({t}\right)=\mathrm{sin}^{βˆ’\mathrm{1}} \left({t}\right)}\\{\mathrm{v}'\left({t}\right)={t}^{\mathrm{7}} }\end{cases}\Rightarrow\begin{cases}{{u}'\left({t}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}βˆ’{t}^{\mathrm{2}} }}}\\{\mathrm{v}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{8}} }\end{cases} \\ $$$$\:\:\:\:\:=\left[{t}^{\mathrm{8}} \mathrm{sin}^{βˆ’\mathrm{1}} \left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{8}} }{\:\sqrt{\mathrm{1}βˆ’{t}^{\mathrm{2}} }}{dt},\:{t}=\mathrm{sin}\vartheta\Rightarrow{dt}=\mathrm{cos}\vartheta{d}\vartheta \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{2}}βˆ’\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} \vartheta{d}\vartheta=\frac{\pi}{\mathrm{2}}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} {xdx}=\frac{\pi}{\mathrm{2}}βˆ’{I}_{\mathrm{1}} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} ={I}_{\mathrm{1}} +\frac{\pi}{\mathrm{2}}βˆ’{I}_{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$
Commented by Ar Brandon last updated on 30/Dec/21
Thanks Sir. Corrected ! Itβ€²s okay now.
$$\mathrm{Thanks}\:\mathrm{Sir}.\:\mathrm{Corrected}\:!\:\mathrm{It}'\mathrm{s}\:\mathrm{okay}\:\mathrm{now}. \\ $$
Commented by Ar Brandon last updated on 29/Dec/21
Commented by Ar Brandon last updated on 29/Dec/21
No, Sir.
$$\mathrm{No},\:\mathrm{Sir}. \\ $$
Commented by mr W last updated on 30/Dec/21
thanks for confirming sir!
$${thanks}\:{for}\:{confirming}\:{sir}! \\ $$
Answered by mr W last updated on 29/Dec/21
we can solve this integral without  calculating it.
$${we}\:{can}\:{solve}\:{this}\:{integral}\:{without} \\ $$$${calculating}\:{it}. \\ $$
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)
$${we}\:{want}\:{to}\:{find}\:{I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \:{with}\: \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} \:{x}\:{dx} \\ $$$$\:{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right)\:{dx} \\ $$$${now}\:{we}\:{apply}\:{method}\:{above}\:{to}\:{find} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right){dx} \\ $$$${y}={f}\left({x}\right)=\mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right)\: \\ $$$$\Rightarrow{x}=\mathrm{sin}^{\mathrm{8}} \:\left({y}\right)={f}^{βˆ’\mathrm{1}} \left({y}\right) \\ $$$${x}\in\left[\mathrm{0},\mathrm{1}\right]\:\Rightarrow{y}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{8}}]{{x}}\right){dx} \\ $$$$\:\:\:\:\:=\mathrm{1}Γ—\frac{\pi}{\mathrm{2}}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} \:\left({y}\right){dy} \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{2}}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{8}} \:\left({x}\right){dx} \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{2}}βˆ’{I}_{\mathrm{1}} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} ={I}_{\mathrm{1}} +\frac{\pi}{\mathrm{2}}βˆ’{I}_{\mathrm{1}} =\frac{\pi}{\mathrm{2}}\:\checkmark \\ $$$${this}\:{works}\:{even}\:{when}\:{we}\:{don}'{t}\:{know} \\ $$$${how}\:{to}\:{calculate}\:{the}\:{concrete}\:{integrals}. \\ $$$$ \\ $$$${for}\:{the}\:{same}\:{reason}\:{we}\:{can}\:{get}\:{e}.{g}. \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{100}} \:{x}\:{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{βˆ’\mathrm{1}} \left(\sqrt[{\mathrm{100}}]{{x}}\right)\:{dx}=\frac{\pi}{\mathrm{2}} \\ $$
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
$${geometrically}\:\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{means}\:{the} \\ $$$${area}\:{under}\:{the}\:{curve}\:{y}={f}\left({x}\right).\:{i}.{e}. \\ $$$$\int_{\mathrm{0}} ^{{a}} {ydx}=\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}={A}_{\mathrm{1}} \\ $$$${we}\:{see}\:{A}_{\mathrm{1}} ={ab}βˆ’{A}_{\mathrm{2}} \:{while}\:{A}_{\mathrm{2}} \:{is}\:{the} \\ $$$${area}\:{under}\:{the}\:{curve}\:{x}={f}^{βˆ’\mathrm{1}} \left({y}\right),\:{i}.{e}. \\ $$$${A}_{\mathrm{2}} =\int_{\mathrm{0}} ^{{b}} {xdy}=\int_{\mathrm{0}} ^{{b}} {f}^{βˆ’\mathrm{1}} \left({y}\right){dy}. \\ $$$${therefore}\:{we}\:{can}\:{get} \\ $$$$\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}={ab}βˆ’\int_{\mathrm{0}} ^{{b}} {f}^{βˆ’\mathrm{1}} \left({y}\right){dy} \\ $$
Commented by Tawa11 last updated on 30/Dec/21
Wow great sir.
$$\mathrm{Wow}\:\mathrm{great}\:\mathrm{sir}. \\ $$

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