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Use-the-reduction-formular-I-n-sin-n-x-dx-1-n-sin-n-1-x-cos-x-n-1-n-I-n-2-to-evaluate-I-n-sin-6-x-dx-

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Question Number 12535 by tawa last updated on 24/Apr/17
Use the reduction formular. I_n = ∫sin^n (x) dx = −(1/n) sin^(n − 1) (x)cos(x) + ((n − 1)/n)I_n − 2 , to evaluate I_(n ) = ∫sin^6 (x) dx
Usethereductionformular.In=∫sinn(x)dx=−1nsinn−1(x)cos(x)+n−1nIn−2,toevaluateIn=∫sin6(x)dx
Answered by mrW1 last updated on 25/Apr/17
I_n = ∫sin^n (x) dx = −(1/n) sin^(n − 1) (x)cos(x) + ((n − 1)/n)I_(n−2) I_6 =∫sin^6 x dx=−(1/6)sin^5 x cos x+(5/6)∫sin^4 x dx I_4 =∫sin^4 x dx=−(1/4)sin^3 x cos x+(3/4)∫sin^2 x dx I_2 =∫sin^2 x dx=−(1/2)sin x cos x+(1/2)∫ dx=(1/2)x−((sin 2x)/4) I_4 =−(1/4)sin^3 x cos x+(3/4)((1/2)x−((sin 2x)/4))=(3/8)x−(3/(16))sin 2x−(1/4)sin^3 cos x I_6 =−(1/6)sin^5 x cos x+(5/6)[(3/8)x−(3/(16))sin 2x−(1/4)sin^3 cos x] =(5/(16))x−(5/(32))sin 2x−(1/(24))sin^3 cos x−(1/6)sin^5 x cos x+C
In=∫sinn(x)dx=−1nsinn−1(x)cos(x)+n−1nIn−2I6=∫sin6xdx=−16sin5xcosx+56∫sin4xdxI4=∫sin4xdx=−14sin3xcosx+34∫sin2xdxI2=∫sin2xdx=−12sinxcosx+12∫dx=12x−sin2x4I4=−14sin3xcosx+34(12x−sin2x4)=38x−316sin2x−14sin3cosxI6=−16sin5xcosx+56[38x−316sin2x−14sin3cosx]=516x−532sin2x−124sin3cosx−16sin5xcosx+C
Commented by tawa last updated on 25/Apr/17
wow, God bless you sir.
wow,Godblessyousir.

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