Help with Integrating this function

[BaconPancakes]

1. For a whole number n ≥0
define

Integrate

Can someone explain how to integrate Tn(x)? Thanks!

 

[Deleted member 4993]

1. For a whole number n ≥0
define

Integrate

Can someone explain how to integrate Tn(x)? Thanks!

Start to use technology....

Using Wolfram-alpha (indefinite integral):


[-n sqrt(1-x^2) sin{n cos^(-1)(x)} -x cos{n cos^(-1)(x)}]/(n^2-1) + constant

This should give you hints for substitution - should you aspire to perform the integration.

 

[DrPhil]

1. For a whole number n ≥0
define

Integrate

Can someone explain how to integrate Tn(x)? Thanks!

I would start with the substitution

\(\displaystyle \theta = \cos^{-1}(x), \; \; x = \cos(\theta), \; \; dx = -\sin(\theta) d\theta\)

Limits of integration, \(\displaystyle x=-1\; \Rightarrow \; \theta=\pi,\; \; \; x=+1\; \Rightarrow \; \theta=0 \)

 

[SAMUELK]

I would start with the substitution

\(\displaystyle \theta = \cos^{-1}(x), \; \; x = \cos(\theta), \; \; dx = -\sin(\theta) d\theta\)

Limits of integration, \(\displaystyle x=-1\; \Rightarrow \; \theta=\pi,\; \; \; x=+1\; \Rightarrow \; \theta=0 \)

and continue:

You have:
cos(n arccos x) dx

Let u = arccos x
x = cos u
dx = - sin u du
cos(n arccos x) dx
= - cos(nu) sin u du
Use a 'product-to-sum' identity

cos A sin B = [sin(A + B) - sin(A - B)]/2

- cos(nu) sin u du = -[sin(nu + u) - sin (nu - u)]/ 2 =

[sin(nu - u) - sin (nu + u)]/ 2 =

[sin((n-1)u) - sin ((n+1)u)]/ 2

I think you can hack it from here. (sorry -- I mean "complete the computation")