# Very Urgent P5 Reduction FormulaWatch

This discussion is closed.
#1
can any one please help me with the following , its very urgent and its on the specification so might come up in the exam

I(n) = int ((sin nx)/(sin x)) n>0

show that ,

I(n+2) = [ 2sin (n+1)x ] + I(n)
-------------
( n+1 )

i need urgent help ..plz can any one do this or give an hint ...plz...
0
#2
sorry (n+1) is the denominator of [2sin (n+1)x ]
0
14 years ago
#3
OK, I thought I was sorted on P5, thanks for pointing this out to me. ARGGGHH - OK, plan B, hope this doesn't come up
0
#4
its also question 12 exercise 3D page 55 the heninmann book P5

plz some one should be able to do it..
0
#5
yeah i was confident but when i checked specs i got scared
0
14 years ago
#6
In+2
= INT [sin(n+1)xcosx + sinxcos(n+1)x]/sinx dx
= INT sin(n+1)x cosx / sinx dx + INT cos(n+1)x dx
= INT sin(n+1)x / sinx d(sinx) + 1/(n+1) sin(n+1)x (then use by parts to deal with the first one)
= sin(n+1)x - INT sinx [(n+1)sinx cos(n+1)x - sin(n+1)x cosx ]/ sin^2x dx + + 1/(n+1) sin(n+1)x
= (n+2)/(n+1) sin(n+1)x - INT [(n+1)sinx cos(n+1)x - sin(n+1)x cosx]/sinx dx
= (n+2)/(n+1) sin(n+1)x - INT [nsinx cos(n+1)x + sin(-nx)]/sinx dx
= (n+2)/(n+1) sin(n+1)x - INT ncos(n+1)x dx + INT sin(nx)/sinx dx
= (n+2)/(n+1) sin(n+1)x - n/(n+1)sin(n+1)x + In
= 2/(n+1) sin(n+1)x + In

Actually it is one of the most difficult integrals I ve ever seen!
0
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