3. Let $a>1$ and $f:\left[\frac{1}{a}, a\right] \rightarrow \mathbb{R}$ be a function with the property that: $f(x)+f\left(\frac{1}{x}\right)=k, \forall x \in\left[\frac{1}{a}, a\right]$. Calculate:
a) $\int_{\frac{1}{a}}^{a} \frac{(x+1) \cdot f(x)}{x \cdot \sqrt{x^{2}+1}} \mathrm{~d} x$;
b) $\int_{\frac{1}{2}}^{2} \frac{(x+1) \cdot \operatorname{arctg} x}{x \cdot \sqrt{x^{2}+1}} \mathrm{~d} x$;

c) Let $a \in\left(0, \frac{\pi}{2}\right)$ and $n \in \mathbb{N}, n \geq 2$. Determine $\lim _{n \rightarrow \infty} \int_{0}^{a} \sqrt[n]{\operatorname{tg} x} d x$.
3. a) The integral is denoted by I, and using the substitution $x=\frac{1}{t}$, we obtain:

$$
\begin{aligned}
& \mathrm{I}=\int_{\frac{1}{a}}^{a} \frac{(t+1) \cdot f\left(\frac{1}{t}\right)}{t \cdot \sqrt{t^{2}+1}} \mathrm{~d} t=\int_{\frac{1}{a}}^{a} \frac{(t+1) \cdot(k-f(t))}{t \cdot \sqrt{t^{2}+1}} \mathrm{~d} t \Rightarrow \\
& \mathrm{I}=\frac{k}{2} \cdot\left(\int_{\frac{1}{a}}^{a} \frac{1}{\sqrt{t^{2}+1}} \mathrm{~d} t+\int_{\frac{1}{a}}^{a} \frac{1}{t \cdot \sqrt{t^{2}+1}} \mathrm{~d} t\right)=\left.k \cdot \ln \left(x+\sqrt{x^{2}+1}\right)\right|_{\frac{1}{a}} ^{a}
\end{aligned}
$$

b) Since $\operatorname{arctg} x+\operatorname{arctg} \frac{1}{x}=\frac{\pi}{2}$, we have $\int_{\frac{1}{2}}^{2} \frac{(x+1) \cdot \operatorname{arctg} x}{x \cdot \sqrt{x^{2}+1}} \mathrm{~d} x=$ $=\left.\frac{\pi}{2} \cdot \ln \left(x+\sqrt{x^{2}+1}\right)\right|_{\frac{1}{2}} ^{2}$.

c) For $\forall x \in\left(0, \frac{\pi}{2}\right), \operatorname{tg} x>x \Rightarrow \int_{0}^{a} \sqrt[n]{\operatorname{tg} x} \mathrm{~d} x>\int_{0}^{a} \sqrt[n]{x} \mathrm{~d} x=$ $=a^{\frac{n+1}{n}}$.

But for $\forall x \in\left(0, \frac{\pi}{2}\right), \sqrt[n]{\operatorname{tg} x}=\sqrt[n]{\underbrace{1 \cdot 1 \cdot \ldots \cdot 1} \operatorname{tg} x} \leq \frac{n-1+\operatorname{tg} x}{n}$.

The required limit is $a$.