Menu Close

please-help-for-3pi-3pi-sin-2009-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 13059 by sandy_suhendra last updated on 12/May/17
please help for ∫_(−3π ) ^( 3π) sin^(2009) x dx
$$\mathrm{please}\:\mathrm{help}\:\mathrm{for} \\ $$$$\int_{−\mathrm{3}\pi\:} ^{\:\:\:\mathrm{3}\pi} \mathrm{sin}^{\mathrm{2009}} \mathrm{x}\:\mathrm{dx} \\ $$
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
$${if}\:{f}\left(−{x}\right)=−{f}\left({x}\right),\:{then} \\ $$$$\int_{−{a}} ^{\mathrm{0}} {f}\left({x}\right){dx}=\int_{−{a}} ^{\mathrm{0}} {f}\left(−{x}\right){d}\left(−{x}\right)=−\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx} \\ $$$$\Rightarrow\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\int_{−{a}} ^{\mathrm{0}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{0} \\ $$$$ \\ $$$${since}\:\mathrm{sin}^{\mathrm{2009}} \:\left(−{x}\right)=−\mathrm{sin}^{\mathrm{2009}} \:{x} \\ $$$$\Rightarrow\int_{−\mathrm{3}\pi} ^{\mathrm{3}\pi} \mathrm{sin}^{\mathrm{2009}} \:{xdx}=\mathrm{0} \\ $$
Commented by sandy_suhendra last updated on 13/May/17
thank′s for your kindness
$$\mathrm{thank}'\mathrm{s}\:\mathrm{for}\:\mathrm{your}\:\mathrm{kindness} \\ $$
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 .
$${I}=\int_{−\mathrm{3}\pi} ^{\mathrm{0}} \mathrm{sin}\:^{\mathrm{2009}} {x}\:{dx}\:+\int_{\mathrm{0}} ^{\mathrm{3}\pi} \mathrm{sin}\:^{\mathrm{2009}} {x}\:{dx} \\ $$$${say}\:\:{I}\:={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${in}\:{the}\:{first}\:{integral}\:\:{let}\: \\ $$$$\boldsymbol{{x}}=−\boldsymbol{{t}}\:\:\:\:\Rightarrow\:\:{dx}=−{dt} \\ $$$${when}\:{x}=−\mathrm{3}\pi\:,\:\:\:\:{t}=\mathrm{3}\pi \\ $$$${and}\:{when}\:\:{x}=\mathrm{0},\:\:{t}=\mathrm{0}\:. \\ $$$${so}\: \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{3}\pi} ^{\mathrm{0}} \mathrm{sin}\:^{\mathrm{2009}} \left(−{t}\right)\left(−{dt}\right) \\ $$$$\:\:\:=\int_{\mathrm{3}\pi} ^{\mathrm{0}} \mathrm{sin}\:^{\mathrm{2009}} {t}\:{dt}\: \\ $$$$\:\:\:=−\int_{\mathrm{0}} ^{\mathrm{3}\pi} \mathrm{sin}\:^{\mathrm{2009}} {t}\:{dt}\: \\ $$$${which}\:{is}\:{equal}\:{to}\:−\boldsymbol{{I}}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =−{I}_{\mathrm{2}} \\ $$$${so}\:\:\boldsymbol{{I}}=\boldsymbol{{I}}_{\mathrm{1}} +\boldsymbol{{I}}_{\mathrm{2}} \:=\mathrm{0}\:. \\ $$
Commented by sandy_suhendra last updated on 13/May/17
thank′s for your kindness
$$\mathrm{thank}'\mathrm{s}\:\mathrm{for}\:\mathrm{your}\:\mathrm{kindness} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *