An expression for the definite integral $I_n = int_0^{pi/4}{tan^n{x},mathrm{d}x}$

I have the following definite integral:

$$I_n = \int_0^{\pi/4}{\tan^n{x}\,\mathrm{d}x}\quad ,\forall n \in \mathbb{N}$$

1. Calculate $I_0$ and $I_1$.



2. Calculate $I_n + I_{n+2}$.


3. Can we deduce $I_n$?

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Here is my solution:

$$I_0 = \int_0^{\pi/4}{dx}=\pi/4$$
$$I_1 = \int_0^{\pi/4}{\tan{x}\,dx}=\int_0^{\pi/4}{\dfrac{\sin{x}}{\cos{x}}\,dx}$$ we put $u = \cos{x} \rightarrow du = -\sin{x}dx$

I found that: $I_1 = \ln{\sqrt{2}}$

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for the second question:

$$I_n+I_{n+2} =\int_0^{\pi/4}{\tan^n{x}\left( 1+\tan^2{x}\right)\,dx}$$ we put $u = \tan{x} \rightarrow du = (1+\tan^2{x})dx$, that leads to:
$$I_n+I_{n+2} = \dfrac{1}{n+1}$$

My question is: Now, can we deduce the expression of $I_n$? I think it will be a recursive relation, Am I right?

Thank you

Answer

You are almost there; just observe that $$I_{n+2} = \frac{1}{n+1} - I_n,$$ so that $$I_n = \frac{1}{n-1} - I_{n-2} = \frac{1}{n-1} - \frac{1}{n-3} + I_{n-4} = \ldots,$$ which we can split into even and odd cases: in the even case, you would obtain $$I_n = \frac{1}{n-1} - \frac{1}{n-3} + \cdots + (-1)^{n/2} I_0,$$ and in the odd case, $$I_n = \frac{1}{n-1} - \frac{1}{n-3} + \cdots + (-1)^{(n-1)/2} I_1.$$