1

Given

$T_n: [-1, 1] \rightarrow \Bbb R$ with $ n \in \Bbb N_0$,

$$ T_n(x) := \cos(n \arccos x)$$

I have to show that

$\displaystyle\int_{-1}^1 T_n(x){1 \over \sqrt{1 - x^2}}\,dx=\pi$ when $n = 0$,

and otherwise it's identical with $0$.

Approach

I began with substituting $x = \sin u \Rightarrow dx = \cos u\,du$. This simplifies the fraction to

$$ {1 \over \sqrt{1 - x^2}}={\cos u \over \sqrt{1 - \sin^2 u}}={\cos u \over \cos u}=1. $$

So, the fraction disappears, and all that is left is:

$$\int_{-1}^1 \cos(n \arccos(\sin u)) du$$

Now, I wondered how this could be simplified again, and my calculator told me that

$$\int_{-1}^1 \cos(n \arccos(\sin u))\,du = \int_{-1}^1 \cos(nu)\,du$$

I didn't understand why this was true, but it was explained here:

Why is $\int \cos(n\arccos(\sin(x)) = \int \cos(nx)$ on $[-1, 1]$?

Now, I don't know where to go from here.

Evaluating the simplified integral gives me:

$$\sin(nx) \over n$$

But when I put in the integration limits, I don't get either of the necessary solutions from above. Where am I mistaken?

Community
  • 1
Julian
  • 1,401

1 Answers1

1

The substitution should be $u=\arccos x$, so $x=\cos u$ and $u\in[0,\pi]$, so $\sqrt{1-\cos^2u}=\sin u$. Then $dx=-\sin u\,du$ and the integral becomes, after substituting the values $\arccos(-1)=\pi$ and $\arccos(1)=0$ to the bounds, $$ \int_{\pi}^0\cos(nu)\frac{1}{\sqrt{1-\cos^2u}}(-\sin u)\,du= \int_0^\pi\cos(nu)\,du= \left[\frac{\sin(nu)}{n}\right]_0^\pi=0 $$ for $n>0$; the case $n=0$ is easy.

egreg
  • 238,574
  • Thank you! But how would I know that? Gilbert Strang recommends to substitute expressions like $a^2 - x^2$ by $x = a \sin u$. – Julian Jul 03 '16 at 09:41
  • @Julian You have $\arccos x$ in the integral; get rid of it! ;-) The difference between substituting $x=a\sin u$ and $x=a\cos u$ is just that in the second case a minus sign sneaks in and it's possible to forget it. – egreg Jul 03 '16 at 09:41
  • How did you conclude $u \in [0, \pi]$? :) – Julian Jul 03 '16 at 09:47
  • Oh, you just put in the values for $\arccos(1)$ and $\arccos(-1)$. – Julian Jul 03 '16 at 09:49
  • @Julian That's the definition of the arc cosine, emphasized to show that, with those values, one can write $\sqrt{1-\cos^2u}=\sin u$. A proper substitution also changes the integration boundaries. – egreg Jul 03 '16 at 09:54
  • But what happens with the $n$ in the denominator within the last fraction on the right side? I'm not allowed to put in $n = 0$. Why does it follow that it's value is $\pi$ then? – Julian Jul 03 '16 at 09:54
  • @Julian When $n=0$, $T_0(x)=1$ is constant. You have $\int_{-1}^{1}\frac{1}{\sqrt{1-x^2}},dx$ that you should be able to solve. – egreg Jul 03 '16 at 09:57