13

This is a problem from MIT integration bee 2017.

$$\int_0^{\pi/2} \frac 1 {1+\tan^{2017} x} \, dx$$

I have tried substitution method, multiplying numerator and denominator with $\sec^2x$, breaking the numerator in terms of linear combination of the denominator and the derivative of it. None of these methods work.

Some hints please?

Community
  • 1
Partha Sarker
  • 547
  • what is MIT integration bee 2017? – Dr. Sonnhard Graubner Nov 13 '17 at 14:28
  • @Dr.SonnhardGraubner http://www.mit.edu/~same/pdf/qualifying_round_2017_answers.pdf – Partha Sarker Nov 13 '17 at 14:30
  • @Dr.SonnhardGraubner Every year, MIT has a competition for undergraduates where they compete to see who can integrate difficult expressions fastest.It's like a spelling bee, but for integrals. – Aaron Nov 13 '17 at 14:30
  • thank you very much, usefull problems – Dr. Sonnhard Graubner Nov 13 '17 at 14:31
  • 3
    The original problem is a definite one which is very easily evaluated. – jonsno Nov 13 '17 at 14:36
  • Perhaps this way of using the word "bee" is not well known outside North America. If it were, I'd say the term "MIT integration bee" is self-explanatory. – Michael Hardy Nov 13 '17 at 14:36
  • Regardless of power it is $\pi/4$ for the case from $0$ to $\pi/2$ – keoxkeox Nov 13 '17 at 14:37
  • 1
    for indefinite case you will have a quite large mess of log(), sin() and cos(), humanly intractable, but doable I suppose – keoxkeox Nov 13 '17 at 14:37
  • 1
    @guyfsone The original question, as posed by the OP, was with an indefinite integral. Is it valid editing for others to change the question so drastically? I understand why it was done of course... but? My comment/question is to be taken at face value - I do not mean to criticize! Comments? – peter a g Nov 13 '17 at 14:44
  • Yes but which is not what is written in the provided link. check his own link. and you see – Guy Fsone Nov 13 '17 at 14:47
  • 1
    @GuyFsone As I said, I quite understand. On the other hand, I think - I know I just said I wasn't criticizing, and now it sounds like I am - sorry!!! - that maybe in this case it would have been better to point it out, and ask the OP to update the question as desired. (I hope I'm not sounding like a jerk.) – peter a g Nov 13 '17 at 14:50

2 Answers2

10

Try using $$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$

and the fact that $\tan(\frac{\pi}{2}-x) = \cot(x)$ to convert to get the following: $$I = \int_0^{\pi / 2} \frac 1 {1+ \tan^{2017}(x)} \, dx = \int_0^{\pi / 2} \frac 1 {1+ \tan^{2017}(\pi/2-x)} \, dx \\ = \int_0^{\pi / 2} \frac 1 {1+ \cot^{2017}(x)} \, dx = \int_0^{\pi / 2} \frac{\tan^{2017}(x)}{1+ \tan^{2017}(x)} \, dx$$

Hence

$$2I = \int_{0}^{\pi / 2} dx = \frac{\pi}{2}$$

Michael Hardy
  • 1
jonsno
  • 7,521
7

Setting the change of variable: $u=\frac\pi2-x $ and since, $\tan x =\cot(\frac\pi2 -x)$ we have, \begin{align} & \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx = \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} (\frac\pi2-u) } \, du \\[10pt] = {} & \int_0^{\frac\pi2}\frac{1}{1+\cot^{2017}u} \, du = \int_0^{\frac\pi2}\frac{\tan^{2017} u}{1+\tan^{2017} u} \, du \color{red}{= \frac{\pi}{2} -\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} u} \, du} \end{align}

That is $$\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx =\frac\pi4$$

Guy Fsone
  • 23,903