∫[(9+X^2)^(1/2)]dx

显示全部楼层 · 2013-3-29 20:56:38

∫ √(9 + x2) dx = ∫ √(32 + x2) dx，于是选用x = 3tanθ，dx = 3sec2θ dθ= ∫ √(9 + 9tan2θ) * 3sec2θ dθ= 9∫ sec3θ dθ= (9/2)[secθtanθ + ln(secθ + tanθ)] + C= (9/2)(x/3)[√(9 + x2)/3] + (9/2)ln[x/3 + √(9 + x2)/3] + C= (1/2)x√(9 + x2) + (9/2)ln[x + √(9 + x2)] + C下面是∫ sec3x...

千问 · 2013-3-29 20:56:38

求不定积分∫[√(9+X2)]dx原式=3∫√[1+(X/3)2)dx 【令x/3=tanu，则x=3tanu，dx=3sec2udu，代入原式得】=9∫sec3udu=9∫secud(tanu)=9[secutanu-∫secu(sec2u-1)du]=9[secutanu-∫sec3...

千问 · 2013-3-29 20:56:38

令x=3tantdx=3sec^2tdt代入原式得原式=∫3sect*3sec^2tdt=9∫sec^3tdt=∫3sect*d(3tant)=9secttant-9∫tantdsect=9secttant-9∫tant*secttantdt=9secttant-9∫sect(sec^2t-1)dt=9secttant...