We evaluate the definite integral of 1/(1+ tan^8(x)) from 0 to pi/2. 00:00 Exploring identities 01:55 Substitution 03:55 Evaluating the integral 07:17 Fun observation
I did it the boring way, multiplying top and bottom of the fraction by (cos x)^8. Then I applied u = pi/2 - x to show the original integral is equal to the integral of (sin x)^8 over the same denominator. Then adding the integral to itself gives the integral from 0 to pi/2 of 1 dx. So the original integral is half that value, which is pi/4.
Clever use to trigonometric identities to simplify the integral greatly! And great insight into the generality of raising tangent to any even power and still obtaining the same result!
This is what i have on my mind at the first sight Let use substitution t = tan(x) Int\limits_{0}^{\infty}\frac{1}{\left(1+t^8 ight)\left(1+t^2 ight)\mbox{d}t} I = Int\limits_{0}^{\infty}\frac{1}{\left(1+t^8 ight)\left(1+t^2 ight)\mbox{d}t} Lets substitute t =\frac{1}{u} Int\limits_{\infty}^{0}\frac{1}{\left(1+\frac{1}{u^8} ight)\left(1+\frac{1}{u^2} ight)\frac{\left(-1 ight)}{u^2}\mbox{d}u}\\ Int\limits_{0}^{\infty}\frac{u^8}{\left(u^8+1 ight)\left(u^2+1 ight)\mbox{d}u}\\ If we add them up we will get 2I = Int\limits_{0}^{\infty}{\frac{1+u^8}{(u^8+1)(u^2+1)}\mbox{d}u} 2I = Int\limits_{0}^{\infty}{\frac{1}{u^2+1}\mbox{d}u} 2I = \frac{π}{2} I = \frac{π}{4} Your calculations are a little bit simpler because I tried to get integal of rational function wich is unnecessary You inspired me to find closed form for coefficients of Chebyshev polynomials but i m stuck in the sum \sum\limits\_{k=m}^{\lfloor\frac{n}{2} floor}{n \choose 2m} \cdot {m \choose k} Mathematica gives the result in terms of binomial coefficient and power of two but this is incorrect for n=0 and i have no access to step by step solution
I had seen a similar problem with sin(x) and cos(x) to pow of half. As mentioned in the end, the special integral interval type is independent of {tan[x]}^(m).
👍In the realm of rigor, some limiting justification is needed at the undefined singularity at π/2 for tan, 0 for cot If f is a function of variable x, requiring Definite Integral between a and b the variable x can be replaced by a + b -x, it appears in this case existence of the Integral can be overlooked.
After completing the substitution, u becomes a mere label inside the equivalent expression. Multiplying I by 2 creates space to include that equivalent as a substitution for one of the terms in the sum, with the obligation to divide the result by 2 at the end.
What's up sir. Its Jay from your first class in furth maths A levels OSA. I'm in my third year in UO bristol now. Looks like your doing well. Best of luck.
i sort of agree.. i think its a poor explanation for whats really going on, which is that for constant bounds, the answer is independent of the variable of integration. kind of like how in an infinite sum, it doesnt matter what you call the index. really the variable of integration is just an index with an infinitesimal step size.
Neat! Can't we just consider tan(x) = 1/cot(x) and then calculate the 2I which gives 1 inside the integral?
Yes, this would simplify the calculations from about 4:15 onwards. Very nice!
I did it the boring way, multiplying top and bottom of the fraction by (cos x)^8. Then I applied u = pi/2 - x to show the original integral is equal to the integral of (sin x)^8 over the same denominator. Then adding the integral to itself gives the integral from 0 to pi/2 of 1 dx. So the original integral is half that value, which is pi/4.
Clever use to trigonometric identities to simplify the integral greatly! And great insight into the generality of raising tangent to any even power and still obtaining the same result!
King's property here to save the day.
Hi Dr. Barker!
Very cool!
This is what i have on my mind at the first sight
Let use substitution t = tan(x)
Int\limits_{0}^{\infty}\frac{1}{\left(1+t^8
ight)\left(1+t^2
ight)\mbox{d}t}
I = Int\limits_{0}^{\infty}\frac{1}{\left(1+t^8
ight)\left(1+t^2
ight)\mbox{d}t}
Lets substitute t =\frac{1}{u}
Int\limits_{\infty}^{0}\frac{1}{\left(1+\frac{1}{u^8}
ight)\left(1+\frac{1}{u^2}
ight)\frac{\left(-1
ight)}{u^2}\mbox{d}u}\\
Int\limits_{0}^{\infty}\frac{u^8}{\left(u^8+1
ight)\left(u^2+1
ight)\mbox{d}u}\\
If we add them up we will get
2I = Int\limits_{0}^{\infty}{\frac{1+u^8}{(u^8+1)(u^2+1)}\mbox{d}u}
2I = Int\limits_{0}^{\infty}{\frac{1}{u^2+1}\mbox{d}u}
2I = \frac{π}{2}
I = \frac{π}{4}
Your calculations are a little bit simpler because I tried to get integal of rational function wich is unnecessary
You inspired me to find closed form for coefficients of Chebyshev polynomials but i m stuck in the sum
\sum\limits\_{k=m}^{\lfloor\frac{n}{2}
floor}{n \choose 2m} \cdot {m \choose k}
Mathematica gives the result in terms of binomial coefficient and power of two but
this is incorrect for n=0 and i have no access to step by step solution
Great
I had seen a similar problem with sin(x) and cos(x) to pow of half. As mentioned in the end, the special integral interval type is independent of {tan[x]}^(m).
👍In the realm of rigor, some limiting justification is needed at the undefined singularity at π/2 for tan, 0 for cot
If f is a function of variable x, requiring Definite Integral between a and b the variable x can be replaced by a + b -x, it appears in this case existence of the Integral can be overlooked.
I think the biggest takeaway here is that all "interesting" trig integrals on youtube evaluate to pi/4.
Why can you change the u's for x at the end and not sub in pi/2 - x which was your u susbstitution?
After completing the substitution, u becomes a mere label inside the equivalent expression. Multiplying I by 2 creates space to include that equivalent as a substitution for one of the terms in the sum, with the obligation to divide the result by 2 at the end.
What's up sir. Its Jay from your first class in furth maths A levels OSA. I'm in my third year in UO bristol now. Looks like your doing well. Best of luck.
I have always disliked the concept "dummy variable".
i sort of agree.. i think its a poor explanation for whats really going on, which is that for constant bounds, the answer is independent of the variable of integration. kind of like how in an infinite sum, it doesnt matter what you call the index. really the variable of integration is just an index with an infinitesimal step size.
@@nathanisbored Good explanation. The variable is like an index variable in a FOR loop in a computer program. Thanks for the reply. 👍
@nathanisbored @Jack_Callcott_AU Good point, and good explanation... thanks for sharing that!
@@nathanisbored Same situation, when a Limit exists, it will not contain the variable(s) used to specify the limit.