MHB Can You Solve This Definite Integral Challenge with Binomial Expansion?

[lfdahl]

Derive an expression for the definite integral:\[I = \int_{0}^{\frac{\pi}{4}}sec^m(x)dx, \;\;\;\;m = 2,4,6,...\]

 

[Opalg]

Derive an expression for the definite integral:\[I = \int_{0}^{\frac{\pi}{4}}sec^m(x)dx, \;\;\;\;m = 2,4,6,...\]

[sp]\[I = \int_0^{\pi/4}\sec^{m-2}x\sec^2x\,dx = \int_0^{\pi/4}\sec^{m-2}x\,d(\tan x) = \int_0^1(1+t^2)^n\,dt,\] where $t = \tan x$ and $n = (m-2)/2.$ The value of this integral is $\dfrac{a(n)}{1\cdot3\cdot5\cdots(2n+1)},$ where $a(n)$ is the $n$th term in Sloane's sequence A076729. This sequence can be expressed in terms of hypergeometric functions, but not in any simpler way.

Another way to express the answer would be to use the binomial theorem to write the integral as \[\int_0^1(1+t^2)^n\,dt = \int_0^1\sum_{k=0}^n{n\choose k} t^{2k}dt = \sum_{k=0}^n{n\choose k}\int_0^1 t^{2k}dt = \sum_{k=0}^n\frac1{2k+1}{n\choose k}.\]
[/sp]

 

[lfdahl]

Hi, Opalg, thankyou for such a detailed and thorough answer!

Spoiler

Yes, I was asking for the solution with binomial expansion

 

[Albert1]

Hi, Opalg, thankyou for such a detailed and thorough answer!

Spoiler

Yes, I was asking for the solution with binomial expansion

Spoiler

Opalg's solution with binomial expansion:
Another way to express the answer would be to use the binomial theorem to write the integral as \[\int_0^1(1+t^2)^n\,dt = \int_0^1\sum_{k=0}^n{n\choose k} t^{2k}dt = \sum_{k=0}^n{n\choose k}\int_0^1 t^{2k}dt = \sum_{k=0}^n\frac1{2k+1}{n\choose k}.\]

Innovative ! I like the solution with binomial expansion