With $n \in \mathbb{N}^*$, consider the function:
$$f : [-1,1] \rightarrow \mathbb{R} \hspace{2cm} f(x) = \cos(n \arccos x)$$
find the volume of the solid of revolution around the $Ox$ axis.
I know that the volume of the solid of revolution around the $Ox$ axis is:
$$V = \pi \int_{-1}^1 (f(x))^2 dx = \pi \int_{-1}^1 \cos^2(n \arccos x) dx$$
But I don't know how to compute that integral. That $n$ inside the cosine complicates things.
Best Answer
Let $t=\arccos x$. Then, $x = \cos t$, $dx= -\sin t dt$ and substitute the integral,
$$V = \pi \int_{-1}^1 \cos^2(n \arccos x) dx$$ $$=-\pi \int_{0}^{\pi}\cos^2(n t) \sin t dt = -\frac\pi2 \int_{0}^{\pi} [1+\cos(2n t)] \sin t dt$$
Use the identity $2\cos a\sin b=\sin(a+b)-\sin(a-b) $, $$ V= -\frac\pi2 \int_{0}^{\pi} \sin t dt - \frac\pi4 \int_{0}^{\pi} [\sin(2n+1)t -\sin(2n-1)t] dt$$ $$ = \pi - \frac\pi4 \frac{\cos(2n+1)\pi - 1}{2n+1} + \frac\pi4 \frac{\cos(2n-1)\pi - 1}{2n-1} $$ $$ = \pi - \frac\pi2 \frac{\cos(2n\pi) + 1}{4n^2-1} = \frac{2\pi (2n^2-1)}{4n^2-1} $$