Question

求解一道定积分计算题

Answer 1

原式=∫<-1,1>[2x²/1+√(1-x²)]dx+∫<-1,1>[xcosx/1+√(1-x²)]dx
=∫<-1,1>[2x²/1+√(1-x²)]dx+0【奇函数在返圆弯对称区间上的漏闷定腔仿积分为零】
=2∫<0,1>[2x²/1+√(1-x²)]dx【偶函数在对称区间上的定积分为其半区间上定积分的2倍】
=4∫<0,1>[x²/1+√(1-x²)]dx
令x=sinα，则dx=cosαdα
且，x=0时，α=0；x=1时，α=π/2
所以，原式=4∫<0,π/2>[sin²α*cosα/(1+cosα)]dα
=4∫<0,π/2>[(1-cos²α)*cosα/(1+cosα)]dα
=4∫<0,π/2>[(1-cosα)*cosα]dα
=4∫<0,π/2>(cosα-cos²α)dα
=4∫<0,π/2>cosαdα-2∫<0,π/2>2cos²αdα
=4sinα|<0,π/2>-∫<0,π/2>(cos2α+1)d(2α)
=4(1-0)-∫<0,π/2>cos2αd(2α)-2∫<0,π/2>dα
=4-sin2α|<0,π/2>-2α|<0,π/2>
=4-(0,0)-2[(π/2)-0]
=4-π

Answer 2

Answer 3

追问

怎么做我明白了，不过最后是4-∏，谢啦

Answer 4

一信唯道定积分滑尺培的困键计算题