# Evaluate the integral integral from 0 to pi/4 of cos(2x)sin(sin(2x)) with respect to x

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$$int_0^pi/4sqrtsin(2x)over cos^2(x) huphikhongdung.vnrm dx=2-sqrt2over picdotGamma^2left(3over 4 ight) ag1$$

My try:

Change $(1)$ to

$$int_0^pi/4sqrt2sec^2(x) an(x) huphikhongdung.vnrm dx ag2$$

$$int_0^pi/4sqrt2 an(x)+2 an^3(x) huphikhongdung.vnrm dx ag3$$

Not sure what substitution lớn use

How may we prove $(1)?$

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By substituting $x=arctan t$ our integral takes the form:

$$I=int_0^1sqrtfrac2t1+t^2,dt$$and by substituting $frac2t1+t^2=u$ we get:$$I = int_0^1left(-1+frac1sqrt1-u^2 ight)fracduu^3/2$$that is straightforward to lớn compute through the substitution $u^2=s$ & Euler"s Beta function:$$I = frac12 left(4+fracsqrtpi ,Gammaleft(-frac14 ight)Gammaleft(frac14 ight) ight).$$The identities $Gamma(z+1)=z,Gamma(z)$ và $Gamma(z)Gamma(1-z)=fracpisin(pi z)$ settle OP"s $(1)$.

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$ewcommandbx<1>,box<8px,border:1px groove navy>displaystyle#1, ewcommandraces<1>leftlbrace,#1, ight brace ewcommandracks<1>leftlbrack,#1, ight brack ewcommanddd huphikhongdung.vnrmd ewcommandds<1>displaystyle#1 ewcommandexpo<1>, huphikhongdung.vnrme^#1, ewcommandic huphikhongdung.vnrmi ewcommandmc<1> huphikhongdung.vncal#1 ewcommandmrm<1> huphikhongdung.vnrm#1 ewcommandpars<1>left(,#1, ight) ewcommandpartiald<3><>fracpartial^#1 #2partial #3^#1 ewcommand oot<2><>,sqrt<#1>,#2,, ewcommand otald<3><>frac huphikhongdung.vnrmd^#1 #2 huphikhongdung.vnrmd #3^#1 ewcommandverts<1>leftvert,#1, ightvert$eginalign&box<#ffe,10px>dsint_0^pi/4 ootsinpars2x over cos^2parsx,dd x =int_0^pi/4 ootsinpars2x over racks1 + cospars2x/2,dd x\<5mm> stackrel2x mapsto x=,,,&int_0^pi/2 ootsinparsx over 1 + cosparsx,dd x=left.Reint_x = 0^pi/2parsz - 1/z over 2ic^1/21 over 1 + parsz + 1/z/2,dd z over ic z, ightvert_ z = expparsic x\<5mm> = &\left. oot2,Imint_x = 0^pi/2pars1 - z^2 over z,ic^1/2dd z over pars1 + z^2, ightvert_ z = expparsic x\<1cm> stackrelmrmas epsilon o 0^+sim &- oot2,Imint_1^epsilonpars1 + y^2 over y^1/2,ic,dd y over pars1 + ic y^2\<3mm> & - oot2,Imint_pi/2^0expparsicrackspi/4 - heta/2 over epsilon^1/2,epsilonexpoic hetaicdd heta\<3mm> &- oot2,Imint_epsilon^1pars1 - x^2 over x^1/2expoicpi/4,dd x over pars1 + x^2\<1cm> stackrelmrmas epsilon o 0^+ o &\root2,int_0^1pars1 + y^2 over y^1/2,1 - y^2 over pars1 + y^2^2,dd y -int_0^1pars1 - x^2 over x^1/2,dd x over pars1 + x^2\<5mm> và =underbrace oot2,int_0^1,1 - y^2 over pars1 + y^2^3/2,dd y over y^1/2_dsmcI_1 -\underbraceint_0^1pars1 - x^1/2 over pars1 + x^3/2,dd x over x^1/2_dsmcI_2 = ,mcI_1 - ,mcI_2label1 ag1endaligneginalignmcI_1 và equiv oot2,int_0^1,1 - y^2 over pars1 + y^2^3/2,dd y over y^1/2 =- oot2,int_0^1,-1/y^2 + 1 over pars1/y + y^3/2,dd y\<5mm> và stackrel1/y + y mapsto y=,,,- oot2int_infty^2dd y over y^3/2implies bx,mcI_1 = 2label2 ag2endaligneginalignmcI_2 và equivint_0^1pars1 - x^1/2 over pars1 + x^3/2,dd x over x^1/2 = int_0^1x^-1/2,pars1 - x^1/2,racks1 - pars-1x^,-3/2,dd x\<5mm> và =mrmBpars1 over 2,3 over 2,_2mrmF_1pars3 over 2,1 over 2;2;-1label3 ag3endalign$dsmrmB$ & $ds,_2mrmF_1,$ are the BetaHypergeometric Functions, respectively. eqref3 is the Euler Type Expression of the Hypergeometric Function.eginalignmboxNote thatquadmrmBpars1 over 2,3 over 2 và =Gammapars1/2Gammapars3/2 over Gammapars2 = pi over 2label4 ag4endalignThe Hypergeometric function is evaluated with theKummer Theorem. Namely,eginalign_2mrmF_1pars3 over 2,1 over 2;2;-1 & =Gammapars2Gammapars7/4 over Gammapars5/2Gammapars5/4 =pars3/4Gammapars3/4 over rackspars1/2pars3/2 ootpirackspars1/4Gammapars1/4\<5mm> và =4,Gammapars3/4 over ootpi,1 over pi/racksGammapars3/4sinparspi/4 =2 over pi, oot2 over piGamma^2pars3 over 4label5 ag5endalignWith eqref4 and eqref5, eqref3 becomes$$bxmcI_2 equivint_0^1pars1 - x^1/2 over pars1 + x^3/2,dd x over x^1/2 = oot2 over piGamma^2pars3 over 4$$such that eqref1 becomes:$$box<20px,#ffe,border:1px dotted navy>ds%int_0^pi/4 ootsinpars2x over cos^2parsx,dd x =2 - oot2 over piGamma^2pars3 over 4$$