Integral of sin^5(x)

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Integral of sin^5(x),
integral of sin^5 x
integral of (sin(x))^5

trig integrals, trigonometric integrals, integral of sin(x), integral of tan(x), integral of sec(x), James Stewart single variable Calculus sect 7.2, integration examples, integral examples, antiderivative examples, u substitution examples, integral practice problems, calculus 2 practice problems, blackpenredpen,

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*Here is a general method to find integral of sin^n(x) where n is a natural number (in the video n=5):*
Let's note U(n) = int(sin^n(x) dx, 0, t) the integral of sinn(x) between 0 and a real t .
I suggest to find a recursion to calculate U(n) using the integration by parts (IP)
_As a remind the IP consists on doing : int(f(x) g'(x) dx, a, b) = [f(b) g(b) - f(a) g(a)] - int(f'(x) g(x) dx, a, b)_
So, U(n+2) = int(sin^(n+2)(x) dx, 0, t)=int(sin^(n) sin²(x)dx, 0, t)
= int(sin^(n)(x) [1-cos²(x)] dx, 0, t)
=
int(sin(n)(x) dx, 0, t) - int(sin(n) cos²(x) dx, 0, t)
= U(n) - int(sin^n(x) cos²(x) dx, 0, t)

Now, we apply the IP to simplify *int(sin^n(x) cos²(x) dx, 0, t)*
To do that, we consider the functions _f(x) = cos(x)_ and _g(x) = sin^(n+1) (x) / (n+1)
_
Then f'(x) = -sin(x) and g'(x) = cos(x) sin^n(x)
Therefore, using the IP, we obtain that :
int(sin^n(x) cos²(x) dx, 0, t)= cos(t) sin^(n+1)(t) /(n+1) - int(- sin^(n+2)(x)/(n+1) dx, 0,t)
= cos(t) sin^(n+1)(t) /(n+1) + U(n+2) / (n+1)
Then, U(n+2) = U(n) - cos(t) sin^(n+1)(t) /(n+1) - U(n+2) / (n+1)
which means that *_U(n+2) = [(n+1)/(n+2)] U(n) - cos(t) sin^(n+1)(t) /(n+2)_* (*)
As we find the recursion (*), let's try to find a close expression of U(n) ; we begin with a pairwise n :
we know U(0) = 0 ; U(2) = U(0)/2 - cos(t) sin(t)/2 = -cos(t) sin(t)/2 ; U(4) = 3U(2)/4 -cos(t) sin^3(t)/4 = -3cos(t) sin(t)/8 - cos(t) sin^3(t)/4 ;...
so, in general, U(2n)= - cos(t) [ sin(t) + sin^3(t)+...+[1/(2n)] sin^(2n-1)(t)
To make it more "beatiful", let's note _a(k) = [fact(2n)/fact(2k)] .[fact(k)/fac(n)]² / 4^[n-k]_
Thus, *U(2n) = - cos(t) . sum( a(k) sin^(2k-1)(t) / (2k) ]* , for k from 1 to n) >> which is calculable by a program if you want to code for example
For odd numbers, U(1) = -cos(t) ; U(3) = 2 U(1)/3 - cos(t) sin²(t)/3 = -cos(t) [ 2/3 - sin²(t)/3] ;..
Let's note _b(k) = [fact(2k+1)/fact(2n+1)] .[fact(n)/fac(k)]² . 4^[n-k]_
In general, *U(2n+1) = - cos(t) . sum ( b(k) sin^(2k) / (2k+1) , for k from 0 to n-1)*

toopytoopy

Maaaan you're handling those pens really good!
Thanks!

basiliasTG

You are my favourite integration teacher! Sometimes I just click on a video for fun:P

shavuklia

Thank you BRPR, I'm studying for my Calc 2 final now and these videos are immensely helpful.

Treegrower

Hi ! This is a lovely tutorial on Integration (using substitution). I would like to know if you have any other video tutorials such as Mechanics (M1 or even M2) and Statistics (S1 or even S2). Your explanation is as clear as crystal.

taslimtorabally

Many many thanks and Love from our Bangladeshi student♥♥♥♥

wahidulislamwasif

No entendí nada de lo que dijiste, pero entendí perfecto la integral (el poder de las matemáticas ;) ) asiático, guapo e inteligente justo mi tipo!! saludos desde Honduras. :*

reynadiaz

You and Julioprofe saved my life, thank you <3

estebanxelhuantzi

thanks, it helps me a lot when my teacher dont want to teach that :(

michaelchristianto

it was quite easy though i messed at the last step.you contents really help a lot.Thanks for keeping up the good work

creativefoxstudio

Unfortunate how i discover this channel the day right before my final... Integrals made so simple

m.chaelx

i would not have gotten this without u fine sir

chrisw

Thank you! You saved my Mathematics exam!

rohanv

Thanks a lot...i was thinking over 1day to solve it at last found it here😇

nayemahmed

i have been thinking about how to integrate 1/[4-6(cosX)^6] , x/[4-6(cosX)^6] or x^2/[4-6(cosX)^6], or in general term x^n / a-bsin^6X as a whole. Thanks.

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