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Reduction formula in integration ...plz help ?
Obtain a reduction formula for :
I(n) = ∫ (from 0 to pi/2) x^n sin x dx .....where n > 1
and hence evaluate I(3).
can anyone plz help me in solving this
3 Answers
- MadhukarLv 79 years agoFavorite Answer
Integrating by parts,
I(n)
= [-x^n cosx] (x=0 to π/2) + n ∫x^(n-1) cosx dx ... (x=0 to π/2)
= 0 + n [ {x^(n-1) sinx} (x=0 to π/2) - (n-1) ∫ x^(n-2) sinx dx ] ... (x=0 to π/2)
= n (π/2)^(n-1) - n(n-1) I(n-2)
=> I(n) + n(n-1) I(n-2) = n (π/2)^(n-1)
I(3)
= 3(π/2)^2 - 6 I(1)
= 3(π/2)^2 - 6 ∫ (from 0 to π/2) x sin x dx ... ( 1 )
∫ (from 0 to π/2) x sin x dx
= - x cosx + ∫ cosx dx ... (x=0 to π/2)
= - x cosx + sinx ... (x=0 to π/2)
= 0 + 1
Plugging in ( 1 ),
I(3) = 3(π/2)^2 - 6.
- Anonymous9 years ago
In tegration by parts twice ---> I_(n) = n(pi/2)^(n-1) - n(n-1)I_(n-2)
---> I_(3) = 3(pi/2)^2 - 6 I_(1)
and I_(1) = 1
---> answer = (3/4)pi^2 - 6
- karpLv 44 years ago
Use integration by skill of aspects on ? (ln x)^n dx to get the help formula: u = (ln x)^n du = n(ln x)^(n-a million) (a million/x) dx dv = dx v = x ? (ln x)^n dx = uv - ? v du ? (ln x)^n dx = x(ln x)^n - n ? (ln x)^(n-a million) dx it would be basic from there.
