Calculate the limitation:$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x $ $$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x
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the answer is 0. This is what I've done so far.
$$ \begin{split}
\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x&=\frac{1}{n}\int_0^{2\pi}{\frac{1}{x+1}}\text{d}\left( \sin nx \right)\\
&=\left. \frac{\sin nx}{n}\cdot \frac{1}{1+x} \right|_{0}^{2\pi}+\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x\\
&=\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x  
\end{split}$$

Following your intuition, an inequality is useful at the point you got to, namely $|\sin(x)| \leq 1$ Using that, we have
$$\left|\frac{1}{n}\int_0^{2\pi} \frac{\sin(nx)}{(1+x)^2}dx\right| \leq\frac{1}{n}\int_0^{2\pi}\frac{1}{(x+1)^2}dx=\frac{1}{n}\frac{-1}{x+1}\bigg|_0^{2\pi}=\frac{1}{n}\left(1-\frac{1}{2\pi+1}\right)<\frac{1}{n}$$
Now letting $n$ go to infinity we get
$$\lim_{n \to \infty} |I| <\lim_{n \to \infty} \frac{1}{n} = 0$$
So $I=0$
