Menu Close

Please-I-need-help-Exercise-We-have-J-n-0-pi-4-tan-n-x-dx-1-Establish-a-recurrence-relation-between-J-n-2-and-J-n-2-Calculate-J-0-and-J-1-then-deduce-the-expression-of-J-n-a




Question Number 105781 by IE last updated on 31/Jul/20
Please, I need help.  Exercise  We have :  J_n = ∫_0 ^( (π/4)) tan^n (x) dx    1) Establish a recurrence relation  between J_(n+2)  and J_n .  2) Calculate J_0  and J_1 , then  deduce the expression of J_n  as a  function of n.  The deduction of the last  question, please.
Please,Ineedhelp.ExerciseWehave:Jn=0π4tann(x)dx1)EstablisharecurrencerelationbetweenJn+2andJn.2)CalculateJ0andJ1,thendeducetheexpressionofJnasafunctionofn.Thedeductionofthelastquestion,please.
Commented by prakash jain last updated on 31/Jul/20
q2088 for n odd
q2088fornodd
Answered by prakash jain last updated on 31/Jul/20
∫_0 ^(π/4) tan^n xdx=∫_0 ^(π/4) tan^(n−2) x(sec^2 x−1)dx  =∫_0 ^(π/4) tan^(n−2) xsec^2 xdx−∫_0 ^(π/4) tan^(n−2) xdx  I_n =[((tan^(n−1) x)/(n−1))]_0 ^(π/4) −I_(n−2)   I_n =(1/(n−1))−I_(n−2)   I_0 =∫_0 ^(π/4) dx=(π/4)  I_1 =∫_0 ^(π/4) tan xdx=((ln 2)/2)  I_2 =1−I_0   I_3 =(1/2)−I_1   I_4 =(1/3)−1+I_0   I_6 =(1/5)−(1/3)+1−I_0   I_(2n) =(−1)^n [(π/4)+Σ_(m=1) ^n  (((−1)^m )/(2m−1))]  Question 2088 for n odd (2n+1)
0π/4tannxdx=0π/4tann2x(sec2x1)dx=0π/4tann2xsec2xdx0π/4tann2xdxIn=[tann1xn1]0π/4In2In=1n1In2I0=0π/4dx=π4I1=0π/4tanxdx=ln22I2=1I0I3=12I1I4=131+I0I6=1513+1I0I2n=(1)n[π4+nm=1(1)m2m1]Question2088fornodd(2n+1)
Commented by IE last updated on 31/Jul/20
thanks sir! I am waiting more  than the answer to the last  question (deduction).
thankssir!Iamwaitingmorethantheanswertothelastquestion(deduction).
Commented by prakash jain last updated on 31/Jul/20
I am not able derive closed  form formula. Summation  notation formula is what i have.  i will continue to think.
Iamnotablederiveclosedformformula.Summationnotationformulaiswhatihave.iwillcontinuetothink.
Commented by prakash jain last updated on 31/Jul/20
Even n formula is above. Odd  n i have mentioned the question  where it is derived  I_(2n+1) =(−1)^n ((1/2)ln2+Σ_(m=1) ^n (((−1)^m )/(2m)))
Evennformulaisabove.OddnihavementionedthequestionwhereitisderivedI2n+1=(1)n(12ln2+nm=1(1)m2m)
Commented by IE last updated on 31/Jul/20
Thanks!
Thanks!
Commented by IE last updated on 31/Jul/20
Thank you very much Sir! Show me  the summation formula, I′m curious  to see what it might look like.
ThankyouverymuchSir!Showmethesummationformula,Imcurioustoseewhatitmightlooklike.
Commented by 1549442205PVT last updated on 01/Aug/20
Why do Sir don′t use the  formular   I_n =(1/(n−1))−I_(n−2 ) to calculate in the case  n is odd like as did forI_5  .Please,sir can  explain ?
WhydoSirdontusetheformularIn=1n1In2tocalculateinthecasenisoddlikeasdidforI5.Please,sircanexplain?

Leave a Reply

Your email address will not be published. Required fields are marked *