SUBIECTUL II Let the sequence $\left(I_{n}\right)_{n \geq 1}$ with the general term:

$$
I_{n}=\int_{0}^{\frac{\pi}{4}} e^{n x}\left(\operatorname{tg}^{n-1} x+\operatorname{tg}^{n} x+\operatorname{tg}^{n+1} x\right) d x, n \geq 1
$$

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Calculate: $\lim _{n \rightarrow \infty} n\left(\sqrt[n^{2}]{n I_{n}}-1\right)$.
## Solution:

We have

$$
\begin{aligned}
& I_{n}=\int_{0}^{\frac{\pi}{4}}\left(e^{x} \tan x\right)^{n-1} e^{x}\left(1+\tan x+\tan^{2} x\right) d x=\ldots \ldots \ldots \\
& =\int_{0}^{\frac{\pi}{4}}\left(e^{x} \tan x\right)^{n-1}\left(e^{x} \tan x\right)^{\prime} x=\left.\frac{1}{n}\left(e^{x} \tan x\right)^{n}\right|_{0} ^{\frac{\pi}{4}}=\frac{1}{n} e^{\frac{n \pi}{4}}
\end{aligned}
$$

Then \( n\left(\sqrt[n^{2}]{n I_{n}}-1\right)=n\left(e^{\frac{\pi}{4 n}}-1\right)=\frac{e^{\frac{\pi}{4 n}}-1}{\frac{\pi}{4 n}} \cdot \frac{\pi}{4} \)

Thus, the required limit is \(\frac{\pi}{4}\). \(\qquad\) (1p)