# 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 $I_n$?

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: $I_1 = \ln{\sqrt{2}}$

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

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, $$I_n = \frac{1}{n-1} - \frac{1}{n-3} + \cdots + (-1)^{(n-1)/2} I_1.$$

• Thank you so much i get it – Navaro Jun 14 '16 at 0:27

Yes.

$$I_{2n} = \frac{1}{2n-1}- \frac{1}{2n-3}+ \cdots + (-1)^{n-1} +(-1)^{n}\pi/4$$

$$I_{2n+1} = \frac{1}{2n}- \frac{1}{2n-2}+ \cdots + (-1)^{n+1}/2 +(-1)^{n}\ln(\sqrt{2})$$

• Thank you but I do not understand your answer, can you explain more? – Navaro Jun 14 '16 at 0:24
• Write $I_2$ in terms of $I_0$, and then $I_4$ in terms of $I_2$ and so on and you will get the even part. Similar for the odd part. – Vaneet Jun 14 '16 at 0:28

By replacing $x$ with $\arctan(t)$ we get:

$$I_n = \int_{0}^{1}\frac{t^n}{1+t^2}\,dt \tag{1}$$ and by writing $t^n$ as $(t^n-t^{n-2})+(t^{n-2}-t^{n-4})+\ldots +A_n$, where $A_n$ is $t^2$ if $n$ is even and $t$ if $n$ is odd, we have: $$I_n = \int_{0}^{1}\left(t^{n-2}+t^{n-4}+\ldots\right)\,dt+\int_{0}^{1}\frac{A_n}{1+t^2}\,dt \tag{2}$$ where the first term gives $\frac{1}{n-1}+\frac{1}{n-3}+\ldots$ and the last one gives $1-\frac{\pi}{4}$ or $\frac{1}{2}\log 2$, depending on the parity of $n$.

• Wow, that's really great! – Navaro Jun 14 '16 at 0:34