Question

求定积分过程

Answer 1

∫(-π/2->π/2) (sinx)^2016 dx
=2∫(0->π/2) (sinx)^2016 dx
=2 [(2015/2016) (2013/2014).....(1/2)(π/2)]
consider
∫(0->π/2) (sinx)^2016 dx
=-∫(0->π/2) (sinx)^2015 dcosx
=-[cosx. (sinx)^2015 ]|(0->π/2) +2015∫(0->π/2) (cosx)^2. (sinx)^2014 dx
=0 +2015∫(0->π/2) (sinx)^2014 dx - 2015∫(0->π/2) (sinx)^2016 dx
(2016) ∫(0->π/2) (sinx)^2016 dx =2015∫(0->π/2) (sinx)^2014 dx
∫(0->π/2) (sinx)^2016 dx
=(2015/2016)∫(0->π/2) (sinx)^2014 dx
=(2015/2016) (2013/2014)∫(0->π/2) (sinx)^2012 dx
=(2015/2016) (2013/2014).....(1/2)∫(0->π/2) dx
=(2015/2016) (2013/2014).....(1/2)(π/2)