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

AI Thread Summary
The discussion focuses on deriving the definite integral \( I = \int_{0}^{\frac{\pi}{4}} \sec^m(x) \, dx \) for even values of \( m \). The integral can be transformed using substitution and integration techniques, resulting in an expression involving Sloane's sequence A076729. Additionally, the binomial theorem is applied to express the integral as a sum, facilitating the calculation of the integral's value. Participants appreciate the innovative approach using binomial expansion. The conversation highlights the complexity and elegance of the mathematical solution.
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Derive an expression for the definite integral:\[I = \int_{0}^{\frac{\pi}{4}}sec^m(x)dx, \;\;\;\;m = 2,4,6,...\]
 
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lfdahl said:
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]
 
Hi, Opalg, thankyou for such a detailed and thorough answer!:cool:

Yes, I was asking for the solution with binomial expansion
 
lfdahl said:
Hi, Opalg, thankyou for such a detailed and thorough answer!:cool:
Yes, I was asking for the solution with binomial expansion
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
 
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