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HARD integration

For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
http://madasmaths.com/archive/maths_...plications.pdf
Original post by runny4
For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
http://madasmaths.com/archive/maths_...plications.pdf


Link does not work for me.
Original post by runny4
For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
http://madasmaths.com/archive/maths_...plications.pdf


Link is disrupted, I can't solve if you can't resolve...
Reply 3
Original post by runny4
For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
http://madasmaths.com/archive/maths_...plications.pdf


Original post by 16Characters....
Link does not work for me.


Original post by Pilate VII
Link is disrupted, I can't solve if you can't resolve...


as this is one of mine but I am a bit busy ...
see if this link works

http://madasmaths.com/archive/maths_booklets/advanced_topics/integration_further_applications.pdf
(edited 8 years ago)
Original post by runny4
For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
http://madasmaths.com/archive/maths_...plications.pdf


For just that part, I'll call it f(u)

f(u)=(u)(r2(u)2)12=u(r2u2)12=f(u)f(-u) = (-u)(r^2-(-u)^2)^\frac{1}{2}=-u(r^2-u^2)^\frac{1}{2}=-f(u)

Hence an odd function.
(edited 8 years ago)
Reply 5
Sub u=tu = -t to see how that term is zero.

I=rrur2u2  du=rrtr2(t)2  dt=rrtr2t2  dt=rrur2u2  du=I.\displaystyle \begin{aligned} I & = \int_{-r}^{r} \frac{u}{\sqrt{r^2-u^2}}\;{du} = \int_{r}^{-r} \frac{t}{\sqrt{r^2-(-t)^2}}\;{dt} \\& = -\int_{-r}^{r} \frac{t}{\sqrt{r^2-t^2}}\;{dt} = -\int_{-r}^{r} \frac{u}{\sqrt{r^2-u^2}}\;{du} \\& = -I. \end{aligned}
Let k=r^2-u^2 ,(dk/-2u)=du,substitute in to the integral and,then easy to integrate and final answer will be r(r-1) or r^2-r
Reply 7
Original post by ghostwalker
For just that part, I'll call it f(u)

f(u)=(u)(r2(u)2)12=u(r2u2)12=f(u)f(-u) = (-u)(r^2-(-u)^2)^\frac{1}{2}=-u(r^2-u^2)^\frac{1}{2}=-f(u)

Hence an odd function.


thanks

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