<s>[INST] Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ I need to find the limit of the sequence :
$$I_n =  \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$
The only thing I have done is : take substitution $\tan x =  z$
this gives 
$$dx = \dfrac{1}{1 + z^2}\,dz$$
So, integral becomes : 
$$I_n = \int_{0}^{\infty} \frac{z^{1/n}}{1 + z^2}\,dz$$ 
After this I am stuck how should I proceed ?
Can someone help me here ?
Thank you.
 [/INST] We first convert the integral into a beta function.
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}}(\tan x)^{\frac{1}{n}} d x &=\int_0^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x \\
&=\frac{1}{2} B\left(\frac{1}{2}+\frac{1}{2 n}, \frac{1}{2}-\frac{1}{2 n}\right) \\
&=\frac{\pi}{2} \csc \left(\frac{1}{2}+\frac{1}{2 n}\right) \pi
\end{aligned}
$$
Then \begin{align}
\lim _{n \rightarrow \infty} I_n=\frac{\pi}{2} \csc \frac{\pi}{2}=\frac{\pi}{2}
\end{align}
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