<s>[INST] Proving the given inequality. (Complex Analysis). We're given a function $P_n(x)$ for $-1\leq x\leq1$ as follows :
$$P_n(x) = \int \limits_0^\pi \dfrac{1}{\pi}(x+i\sqrt{1-x^2} \cos\theta)^n \, d\theta$$
for $n=(0,1,2,3,\ldots)$, we need to prove that $|P_n(x)| \leq 1$.
I tried the following :
Let $z=x+i\sqrt{1-x^2}\cos\theta$ 
I somehow want to prove that $|z|=|x+i\sqrt{1-x^2}\cos\theta|\leq1$, as that would imply that $|z^n|\leq1$, as $|z^n|=|z|^n$.
$$|z| = \sqrt{x^2+(1-x^2)\cos^2 \theta} \Longrightarrow \sqrt{x^2 \sin^2\theta + \cos^2\theta}.$$
Now, $x^2\leq1$ and $\sin^2\theta \leq1 \Longrightarrow x^2\sin^2\theta \leq1$.
Also, $\cos^2\theta \leq1 \Longrightarrow \sqrt{x^2\sin^2\theta+\cos^2\theta} \leq \sqrt{2}$, 
But " $\leq1$" condition is required...
That's the point where I am stuck, could anyone help me with this?
 [/INST] \begin{align}
x^2\sin^2 t+\cos^2 t
&=x^2(1-\cos^2 t+\cos^2 t\\
&=(1-x^2)\cos^2t+x^2\\
\text{so}\\
|x^2\sin^2t+\cos^2t|
&\le|(1-x^2)\cos^2t+x^2|\\
&\le|(1-x^2)\cos^2t|+|x^2|\\
&\le|1-x^2|+|x^2|\\
&\le(1-x^2)+x^2\\
&=1
\end{align}
</s>