maths question (1 Viewer)

[omniscience]

1. Find the sum of the geometric series 1-t^2+t^4-t^6+...+t^(4n) and hence show that for 0<T<X,< b>
1/(1+t^2) < 1-t^2+t^4-t^6+...+t^(4n)

2. Find 1-t^2 + t^4 - t^6 + ...+ t^(4n) - t^(4n+2) and hnce show that for 0 < t < x
1-t^2+t^4-t^6 + ...+ t^(4n) < 1/(1+t^2) + t^(4n+2)

Help me with these questions.

edit: dw, got them.

Can anyone help me with these then?
int (1/(sqrt(x)*(1+x))) dx

AND int (1/(e^(-x) + e^x)) dx

Use the inverse chain rule to find the answer

 

[clintmyster]

for the second one

int (1/(e^(-x) + e^x)) dx


-------------------------------

multiply numerator and denominator by e^x
=int e^xdx / (1 + e^2x )

now you look at the table of standard integrals and you use the tan^-1 one. I'l take out e^x as it is a constant.

= e^x int dx / (1 + (e^x )²)
= e^x x 1/e^x x tan^-1 (e^x /1) + C
= tan^-1 (e^x) + C


and as stupid as this sounds, I forget how to do the first one the 3u way but il try and remember. I got an answer for it 2tan^-1(rootx) + C. Tell me if thats right if you have answers please.

 

[azureus88]

yeh i got clintmysters answer as well

let x = u^2
dx = 2udu
∫ 2u/[u(1+u^2)]du
= 2∫ 1/(1+u^2)du
= 2 tan<SUP>-1</SUP>(u) + c
=2 tan<SUP>-1</SUP>(sqrtx) + c

btw, clintmyster, i dont think u can take out e^x as its not a constant.

so its:
let u = (e^x)
du=(e^x)dx
∫ e^x/[1+e^2x)]du
=∫ 1/[1+u^2)]dx
= tan-1 (u) + c
= tan-1 (e^x) + c

 

[clintmyster]

yeh i got clintmysters answer as well

let x = u^2
dx = 2udu
∫ 2u/[u(1+u^2)]du
= 2∫ 1/(1+u^2)du
= 2 tan^-1(u) + c
=2 tan^-1(sqrtx) + c

btw, clintmyster, i dont think u can take out e^x as its not a constant.

so its:
let u = (e^x)
du=(e^x)dx
∫ e^x/[1+e^2x)]du
=∫ 1/[1+u^2)]dx
= tan-1 (u) + c
= tan-1 (e^x) + c

yeah your right. I actually screwed that question up so OP dont look at my method.