Connecting these Two Integrals with Integration by Parts

Integration by Parts challenge problem, integral of tan^12(x)*sec(x) from 0 to pi/4, Connecting these Two Integrals with Integration by Parts

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blackpenredpen | 曹老師

[yaleng]

Integral addiction 6: Put down a +C for definite integrals

[wih]

no + C because it's a definite integral and k is a constant

[shanmugasundaram]

We can calculate K by reducing the power successively from 10, 8, 6, 4, 2 and following the same method repeatedly.The answer is quiet complicated with alternating terms.

[crtwrght]

I ended up doing this slightly differently, considering the function as tan^11(x) . sec(x)tan(x) and then picking sec(x)tan(x) as my "I" in integration by parts, since it just goes to sec(x). Obviously it then proceeds very similarly, second integral gets a sec^2(x) from the derivative of tan^11(x) which we express as 1+tan^2(x) and split into two integrals, one of tan^10(x)sec(x) (i.e. a multiple of k) and one in tan^12(x)sec(x) (i.e. the original equation. Whole thing quickly becomes12(Integral)=sqrt(2) - 11k
Cool problem.

[garyhuntress]

The vids where you algebraically manipulate an entire integral are really interesting!

[angelmendez-rivera]

A derivation of a general case. Suppose the integral of [tan(t)^n]sec(t) from t = 0 to t = π/4 is equal to S. Then to calculate the integral on the same interval of [tan(t)^(n + 2)]sec(t), simply notice that the integrand is equivalent to [tan(t)^n][tan(t)^2]sec(t) = [tan(t)^n][sec(t)^3] - [tan(t)^n][sec(t)]. The given integration of this is the integral over the same interval of [tan(t)^n][sec(t)^3], minus S. The remaining integral can be performed by using integration by parts, integrating [tan(t)^n][sec(t)^2] and differentiating sec(t). The integration results in [tan(t)^(n + 1)]/(n + 1) and the differentiation in sec(t)tan(t). Then this integral is equivalent to [tan(t)^(n + 1)]sec(t)/(n + 1) (evaluated at boundaries) - integral from t = 0 to t = π/4 of [tan(t)^(n + 2)]sec(t)/(n + 1). If the desired integral is T, then T = sqrt(2)/(n + 1) - T/(n + 1) - S by the above equation in conjunction with the first. Then [(n + 2)/(n + 1)]T = sqrt(2)/(n + 1) - S. This implies T = sqrt(2)/(n + 2) - [(n + 1)/(n + 2)]S. This is the general case formula. In fact, this can just be made more general to T = C/(n + 2) - [(n + 1)/(n + 2)]S if we let our integration bounds be t = a and t = b. Then C = [tan(b)^(n + 1)]sec(b) - [tan(a)^(n + 1)]sec(a).

[wibufisika]

Another video from Black Pen Red Pen.

[menukasharma]

no +c because the solutio will be a ligitimate number as the integral has bounds

[anishadj]

Limit as n goes to infinity of the integral from 0to1 of (ln x)^n

[GaryTugan]

So I stopped the vid and spent 20 mins.... found the same solution but with less steps and complexity! Wondering.... how can I best share this with you? (Note: I STARTED with the DI method ) 👍

[Sitanshu_Chaudhary]

Can you integrate 0 to pi/2 (sin(nx))^2/(sinx)^2 please

[keithmasumoto]

Can you please explain how the Riemann zeta function, expressed as an infinite product, could possibly be zero when all of its factors are of the form 1/s. thank you!

[angelmendez-rivera]

No, you need not add an integration constant +C to your boxed answer, because the answer is a definite integral, and thus a number, not a set of a functions as it would be if it were instead an indefinite integral.

[BigDBrian]

K already contains a +C

edit: nvm I'm dumb it's just a definite integral so both k and the answer are just a number, not a function

[smithsmithsmith]

Hey BPRP, why don't you find the length of arcsin x from -1 to 1?) This is actually a little challenging.

[SovietOnion]

You should solve the cool equation a/(b+c) + b/(a+c) + c/(a+b) = 4

[neilgerace]

9:12 I don't know how I can cope with that much work :)

[srpenguinbr]

differentiate tan^11 and integrate tanxsecx badabum badabin

[yaleng]

Wait, 12th power? NO, no, no, no, this is totally not ok.fml

[snejpu]

Do not write +C, because we have a definite value for the integral.