How to compute $I_n= \int_{0}^{\pi/4} x\tan^{2n}x\, \mathrm{d}x $

Question

Consider the following integral

$$I_n= \int_{0}^{\pi/4} x\tan^{2n}x \, \mathrm{d}x $$

Then compute $I_n + I_{n+1}$ as a function of $n$ $\forall n\geq0$.

This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. — user37238, Mar 17 '16 at 11:29

score 5 · Answer 1 · answered Mar 17 '16 at 11:35

5

Hint: $$ \tan^{2n}(x)+\tan^{2n+2}(x)=\tan^{2n}(x)\sec^2(x) $$ and $$ \frac{\mathrm{d}}{\mathrm{d}x}\tan(x)=\sec^2(x) $$ Try integrating by parts.

answered Mar 17 '16 at 11:35

robjohn

345,667

I have to compute $I_n + I_{n+1}$ as a fonction of $n$. We have $I_n + I_{n+1}= \frac{\pi}{4(2n+1)} + \frac{1}{2n+1} \int_{0}^{\pi/4} tan^{2n+1}(x) , \mathrm{d}x $ . How can i continue ? – John Mar 17 '16 at 12:25
@John: Take a look at my answer. :) – Hosein Rahnama Mar 17 '16 at 12:34
@H. R. i understand , thanks – John Mar 17 '16 at 12:51
given $\pi/4 > \epsilon > 0$, How can we show that : $$\int_0^{\pi/4} \tan^n x\ dx < (\pi/4 - \epsilon) \tan(\pi/4 - \epsilon)^n + \epsilon $$ ? – John Mar 17 '16 at 13:19

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

First, notice that

$$\begin{align} I_n + I_{n+1} &= \int_{0}^{\pi/4} x \left( \tan^{2n}x +\tan^{2n+2}x \right) \, \mathrm{d}x \\ &= \int_{0}^{\pi/4} x\tan^{2n}x \left(1+ \tan^2 x\right) \, \mathrm{d}x \\ &= \int_{0}^{\pi/4} x\tan^{2n}x \, \sec^2 x \, \mathrm{d}x \\ &= \dfrac{1}{2n+1} x \tan^{2n+1} x |_{0}^{\pi/4} - \dfrac{1}{2n+1} \int_{0}^{\pi/4} \tan^{2n+1}x \, \mathrm{d}x \\ &= \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - \int_{0}^{\pi/4} \tan^{2n+1}x \, \mathrm{d}x \right) \\ &\equiv \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - J_n \right) \end{align}$$

and for computing $J_n$ I refer you to this post. For the sake of completeness, I just mention the final result

$$\begin{align} J_n &=\int_{0}^{\pi/4} \tan x ^{2n+1} \mathrm dx \\ &= \left[ \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}\tan x ^{2(n-k)}+(-1)^{n+1}\ln \cos x \right]_{0}^{\pi/4} \\ &= \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}+\frac{(-1)^n}{2}\ln2 \end{align}$$

and hence

$$\boxed{ I_n + I_{n+1} = \dfrac{1}{2n+1} \left( \dfrac{\pi}{4} - \sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}-\frac{(-1)^n}{2}\ln2 \right) }$$

edited Apr 13 '17 at 12:21

Community

1

answered Mar 17 '16 at 12:28

Hosein Rahnama

14,882

what about the convergence of $(I_n)_{n \in N}$ ? – John Mar 17 '16 at 12:58
@John Do you mean $\lim_{n \to \infty} I_n$ exist or not? If so, I think it is better to look at $I_n= \int_{0}^{\pi/4} x\tan^{2n}x, \mathrm{d}x$. – Hosein Rahnama Mar 17 '16 at 13:02
yes i mean $\lim I_n$ – John Mar 17 '16 at 13:03
@John: Take a look at this post. – Hosein Rahnama Mar 17 '16 at 13:06
i understand, thank you – John Mar 17 '16 at 13:08

How to compute $I_n= \int_{0}^{\pi/4} x\tan^{2n}x\, \mathrm{d}x $

2 Answers2