∫cost+2dt

惠矿变2549高数求∫(t - sint)^2sintdt的积分?
邴聪苑18624649440 ______ 解法如下: ∫(t-sint)^2sintdt =∫(t^2sint+sint^2sint-2tsint^2)dt =∫t^2sintdt+∫(1-cost^2)sintdt-2∫tsint^2dt =-∫t^2dcost-∫(1-cost^2)dcost-∫t*(1-cos2t)dt =-t^2cost+∫2costdt-cost+cost^3/3-t^2/2+∫tdsin2t =-t^2cost+2sint-cost+cost^3/3-t^2/2+tsin2t-∫sin2tdt =-t^2cost+2sint-cost+cost^3/3-t^2/2+tsin2t-1/2*sin2t^2

惠矿变2549(x3+1)/(x2+1)2 的积分 -
邴聪苑18624649440 ______ 令x=tant,dx=(sect)2dt,原式=∫(tant)3+1/(sect)2dt=∫tant(sint)2dt+∫(cost)2dt=∫tantdt-∫costsintdt+t/2+sin2t/4+c=-ln|cost|-∫sin2td(2t)/4+t/2+sin2t/4+c=-ln1/√1+x2+(cos2t+sin2t)/4+(arctanx)/2+c=(ln(1+x2)+(x+1)/(1+x2)+arctanx)/2+c

惠矿变2549求教根号下1 - x^2的积分 -
邴聪苑18624649440 ______ ∫√1-x^2 dx 设x=sint 那么 dx=d(sint)=costdt ∫√1-x^2 dx =∫cost*costdt=∫cost^2dt =0.5*∫(1+cos2t)dt =0.5*[∫dt+∫cos2tdt] =0.5*[t+∫cos2td2t /2] =0.5*[t+sin2t /2]+C t=arcsinx 原式 arcsinx /2 +sint*cost /2 +C cost=√1-sint^2=√1-x^2 原式 arcsinx /2 +x*√1-x^2 /2 +C

惠矿变2549∫x^3*(√1 - x^2)dx -
邴聪苑18624649440 ______ ^^^解法1:令x=sin t,则 ∫x^3√(1-x^2)dx =∫(sin t)^3*(cost)^2 dt =-∫[1-(cost)^2]*(cost)^2 d cost =-(cost)^3/3+(cost)^5/5+C =-(1-x^2)*√(1-x^2)/3+(1-x^2)^2*(√1-x^2)/5 +C 解法2:∫x^3√(1-x^2)dx =1/2*∫[(1-x^2)-1]√(1-x^2)d(1-x^2) =1/2*[2/5(1-x^2)^(5/2)-2/3(1-x^2)^(3/2)]+C =1/5(1-x^2)^(5/2)-1/3(1-x^2)^(3/2)+C

惠矿变2549根号下(4 - X*X)被积的一个原函数是?
邴聪苑18624649440 ______ F(x)=∫√(4-x^2)dx=2∫√[1-(x/2)^2]dx |x/2|≤1, 令sint=x/2,则x=2sint F(x)=2∫√(1-sint^2)d(2sint)=4∫cost^2dt =4∫costd(sint)=4costsint-4∫sintd(cost)=4costsint+4∫sint^2dt =4costsint+4∫(1-cost^2)dt=4costsint+4t-4∫cost^2dt =4costsint+4t-F(x) F(x)=2costsint+2t+C=sin2t+2t+C =sin[2arcsin(x/2)]+2arcsin(x/2)+C

惠矿变2549∫costˆ2dt等于多少 -
邴聪苑18624649440 ______ ∫(tant)∧2dt=∫(secx)^2-1dt=tant-t+c ∫[(cost)∧2]/[(sint)∧4]dt=-∫(cott)^2dcott=-1/3(cott)^3+c.

惠矿变2549√(1 ㎡)的原函数是多少 -
邴聪苑18624649440 ______ 设m=sint,则:∫√(1-m^2)dm=∫√(1-sin^2t)dsint=∫cost*costdt=∫(1+cos2t)/2dt=(1/2)∫dt+(1/4)∫cos2td2t=t/2+(1/4)sin2t+c.=(1/2)arcsinm+(1/2)m√(1-m^2)+c.

惠矿变2549根号(a^2 - x^2)dx的形式怎么求? -
邴聪苑18624649440 ______ 令x=asint,t=arcsinx/a 原式=∫√[a^2(1-sint·sint)]acostdt =a^2∫(cost)^2dt =a^2∫1/2+1/2cost/2dt =a^2t/2+∫cott/2dt/2 =a^2t/2+sint/2+c =(a^2arcsinx/a)/2+sin[(arcsinx/a)/2]+c

惠矿变2549cost的平方的原函数是?求解答 -
邴聪苑18624649440 ______ =∫(cost)^2dt=∫(1+cos2t)/2dt=t/2+sin2t/4+C其中C为常数t/2+sin2t/4+C都是它的原函数!!!

惠矿变2549根号a^2+x^2的积分公式是什么? -
邴聪苑18624649440 ______ 解: ∫√(a^2-x^2)dx 设x=asint 则dx=dasint=acostdt a^2-x^2 =a^2-a^2sint^2 =a^2cost^2 ∫√(a^2-x^2)dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫(cos2t+1)/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t) 将x=asint代回 ∫√(a^2-x^2)dx=x√(a^2-x^2)/2+a^2*...