Integration (2 Viewers)

Petyo · May 19, 2009

Can someone please tell me if the integral S 1/ (5-4cosx) dx from x=0 to x= pi can be calculated??

tommykins · May 19, 2009

let t = tan(@/2) and do it from there.

remeber to change your limits

Petyo · May 19, 2009

but you can't have tan(pi/2) can you?

Petyo · May 19, 2009

the full question is :
If U(n) = S (cosnx)/ (5-4cosx) dx, prove that U(n+1) + U(n-1) - (5/2)U (n) = 0. then find U (0).
and I couldn't do the second part...

Timothy.Siu · May 19, 2009

Petyo said:
the full question is :
If U(n) = S (cosnx)/ (5-4cosx) dx, prove that U(n+1) + U(n-1) - (5/2)U (n) = 0. then find U (0).
and I couldn't do the second part...

well,

U (0)=2/5 (U(n+1) + U(n-1))

Petyo · May 20, 2009

I know...

Petyo · May 20, 2009

YOu mean U(0) = 2/5 ( U(1) + U(-1) )??

Timothy.Siu · May 20, 2009

Petyo said:
YOu mean U(0) = 2/5 ( U(1) + U(-1) )??

precisely,
is it cos^n x or just cos nx

Petyo · May 20, 2009

cos nx, I've asked my teacher this question as well = =

Timothy.Siu · May 20, 2009

Petyo said:
cos nx, I've asked my teacher this question as well = =

ok, well u know cos -x=cos x
which doesn't really help lol

omniscience · May 20, 2009

i still personally reckon 1 - 1 = 0

don't u think?

Petyo · May 20, 2009

I tried applying the equation proven in the first part and simplified and obtained something which looked like U(0) = U(0) --> 0=0 @_@

study-freak · May 20, 2009

$LHS=U(n+1) + U(n-1) - \frac{5}{2}U (n) \\\because $We are given that \int_{0}^{\pi}\frac{cosnx}{5-4cosx}dx, \\$ LHS=\int_{0}^{\pi}\frac{cos(n+1)x+cox(n-1)x- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{2coxnxcosx- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{-\frac{1}{2}cosnx(5-4cosx)}{5-4cosx}dx \\=-\frac{1}{2}\int_{0}^{\pi}cosnxdx \\==-\frac{1}{2n}\left [ sinnx \right ]_{0}^{\pi} \\=0 \\=RHS \\Q.E.D. \\Now, U(0) \\=\int_{0}^{\pi}\frac{1}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{2}{5-4cosx}dx \\=\frac{2}{5}[U(1)+U(-1)] \\=\int_{0}^{\pi}\frac{cosx+cos(-x)}{5-4cosx}dx \\=2\int_{0}^{\pi}\frac{cosx}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{4cosx}{5-4cosx}dx \\\therefore \frac{1}{2}\int_{0}^{\pi}\frac{2-4cosx}{5-4cosx}dx=0 \\\frac{1}{2}\int_{0}^{\pi}(1-\frac{3}{5-4cosx})dx=0 \\\frac{1}{2}(\pi-3U(0))=0 \\U(0)=\frac{\pi}{3}$
Took a long time to do this... It's an interesting question though.
Where did you get this question?

Drongoski · May 20, 2009

Great effort study-freak!

Per Timothu.Siu's suggestion, for U(0):

$\100dpi Let \ \ tan\frac{x}{2}\ =\ t \Rightarrow cosx\ =\ \frac{1-t^{2}}{1+t^{2}};\ \ dx\ =\ \frac{2dt}{1+t^{2}}\\ \\ U_{0}\ =\ \int_{0}^{\pi }\frac{1}{5-4cosx}dx\ =\ \int_{x=0}^{\ x=pi } \frac{1}{5\ -\ 4(\frac{1-t^{2}}{1+t^{2}})}\times \frac{2dt}{1+t^{2}}\\ \\ =\ \ \int_{x=0}^{\ x=pi }\frac{2}{1+9t^{2}}dt\ \ =\ \ \frac{2}{3}tan^{-1}(3t)]^{\ x=pi }_{x=0}\ \ =\ \ \left [ \frac{2}{3}tan^{-1}(3tan\frac{x}{2}) \right ]^{\pi }_{0}\\ \\ =\ \ \frac{2}{3}[tan^{-1}(3\ tan\frac{\pi }{2})\ -\ tan^{-1}(3tan\ 0)]\ \ =\ \ \frac{2}{3}[\frac{\pi }{2}\ -\ 0]\ \ =\ \ \frac{\pi }{3}$

Edit

Solution amended.

study-freak · May 20, 2009

Yea, that works nicely but the question says "then find U(0)" or something similar so I thought I should use the previous part of the question.

Petyo · May 20, 2009

study-freak said:
$LHS=U(n+1) + U(n-1) - \frac{5}{2}U (n) \\\because $We are given that \int_{0}^{\pi}\frac{cosnx}{5-4cosx}dx, \\$ LHS=\int_{0}^{\pi}\frac{cos(n+1)x+cox(n-1)x- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{2coxnxcosx- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{-\frac{1}{2}cosnx(5-4cosx)}{5-4cosx}dx \\=-\frac{1}{2}\int_{0}^{\pi}cosnxdx \\==-\frac{1}{2n}\left [ sinnx \right ]_{0}^{\pi} \\=0 \\=RHS \\Q.E.D. \\Now, U(0) \\=\int_{0}^{\pi}\frac{1}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{2}{5-4cosx}dx \\=\frac{2}{5}[U(1)+U(-1)] \\=\int_{0}^{\pi}\frac{cosx+cos(-x)}{5-4cosx}dx \\=2\int_{0}^{\pi}\frac{cosx}{5-4cosx}dx \\=\frac{1}{2}\int_{0}^{\pi}\frac{4cosx}{5-4cosx}dx \\\therefore \frac{1}{2}\int_{0}^{\pi}\frac{2-4cosx}{5-4cosx}dx=0 \\\frac{1}{2}\int_{0}^{\pi}(1-\frac{3}{5-4cosx})dx=0 \\\frac{1}{2}(\pi-3U(0))=0 \\U(0)=\frac{\pi}{3}$
Took a long time to do this... It's an interesting question though.
Where did you get this question?

wow nice work.

I didn't think of moving 1/2 S 4cosx / ( 5- 4cosx) dx onto the other side like what you did in step 6 of the second part of the question...
My teacher told me that he got this question from last year trial @@

Petyo · May 20, 2009

Drongoski said:
Great effort study-freak!

Per Timothu.Siu's suggestion, for U(0):

$\100dpi Let \ \ tan\frac{x}{2}\ =\ t \Rightarrow cosx\ =\ \frac{1-t^{2}}{1+t^{2}};\ \ dx\ =\ \frac{2dt}{1+t^{2}}\\ \\ U_{0}\ =\ \int_{0}^{\pi }\frac{1}{5-4cosx}dx\ =\ \int_{0}^{\pi } \frac{1}{5\ -\ 4(\frac{1-t^{2}}{1+t^{2}})}\times \frac{2dt}{1+t^{2}}\\ \\ =\ \ \int_{0}^{\pi }\frac{2}{1+9t^{2}}dt\ \ =\ \ \frac{2}{3}tan^{-1}(3t)]^{\pi }_{0}\ \ =\ \ \left [ \frac{2}{3}tan^{-1}(3tan\frac{x}{2}) \right ]^{\pi }_{0}\\ \\ =\ \ \frac{2}{3}[tan^{-1}(3\ tan\frac{\pi }{2})\ -\ tan^{-1}(3tan\ 0)]\ \ =\ \ \frac{2}{3}[\frac{\pi }{2}\ -\ 0]\ \ =\ \ \frac{\pi }{3}$

um... this is really dumb of me but I quite don't understand how
$tan^{-1}(3\ tan\frac{\pi }{2}) equals \frac{\pi }{2}$

Petyo · May 20, 2009

btw thanks so much people for helping me :X:X:X

shaon0 · May 20, 2009

Petyo said:
um... this is really dumb of me but I quite don't understand how
$tan^{-1}(3\ tan\frac{\pi }{2}) equals \frac{\pi }{2}$

I think they took the limit as x -> pi/2. lim {x -> (pi/2)} arctan(3tan(x))=pi/2.

study-freak · May 20, 2009

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