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Question Number 13059 by sandy_suhendra last updated on 12/May/17

please help for ∫_(−3π ) ^( 3π) sin^(2009) x dx

pleasehelpfor∫−3π3πsin2009xdx

Answered by mrW1 last updated on 12/May/17

if f(−x)=−f(x), then ∫_(−a) ^0 f(x)dx=∫_(−a) ^0 f(−x)d(−x)=−∫_0 ^a f(x)dx ⇒∫_(−a) ^a f(x)dx=∫_(−a) ^0 f(x)dx+∫_0 ^a f(x)dx =−∫_0 ^a f(x)dx+∫_0 ^a f(x)dx=0 since sin^(2009) (−x)=−sin^(2009) x ⇒∫_(−3π) ^(3π) sin^(2009) xdx=0

iff(−x)=−f(x),then∫−a0f(x)dx=∫−a0f(−x)d(−x)=−∫0af(x)dx⇒∫−aaf(x)dx=∫−a0f(x)dx+∫0af(x)dx=−∫0af(x)dx+∫0af(x)dx=0sincesin2009(−x)=−sin2009x⇒∫−3π3πsin2009xdx=0

Commented by sandy_suhendra last updated on 13/May/17

thank′s for your kindness

thank′sforyourkindness

Answered by ajfour last updated on 12/May/17

I=∫_(−3π) ^0 sin^(2009) x dx +∫_0 ^(3π) sin^(2009) x dx say I =I_1 +I_2 in the first integral let x=−t ⇒ dx=−dt when x=−3π , t=3π and when x=0, t=0 . so I_1 =∫_(3π) ^0 sin^(2009) (−t)(−dt) =∫_(3π) ^0 sin^(2009) t dt =−∫_0 ^(3π) sin^(2009) t dt which is equal to −I_2 I_1 =−I_2 so I=I_1 +I_2 =0 .

I=∫−3π0sin2009xdx+∫03πsin2009xdxsayI=I1+I2inthefirstintegralletx=−t⇒dx=−dtwhenx=−3π,t=3πandwhenx=0,t=0.soI1=∫3π0sin2009(−t)(−dt)=∫3π0sin2009tdt=−∫03πsin2009tdtwhichisequalto−I2I1=−I2soI=I1+I2=0.

Commented by sandy_suhendra last updated on 13/May/17

thank′s for your kindness

thank′sforyourkindness

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