Question

高数题求详细解答

高数题求详细解答第四题

Answer 1

用到第二个重要极限

Answer 2

(1)令√x=t,则x=t²,dx=2tdt
原式=2∫tsintdt
=-2∫td(cost)
=-2(tcost-∫costdt)
=-2(tcost-sint)+C
=-2√xcos√x-2sin√x+C
(2)原式=1/2*∫arcsinxd(x²)
=1/2*[x²arcsinx-∫x²d(arcsinx)]
=1/2*[x²arcsinx-∫x²dx/√(1-x²)]~~~~①
令x=sint,t∈(-π/2,π/2),则t=arcsinx,√(1-x²)=cost,dx=costdt
∫x²dx/√(1-x²)
=∫sin²t*costdt/cost
=∫sin²tdt
=∫(1-cos2t)/2*dt
=1/2*∫dt-1/4*∫cos2td(2t)
=t/2-sintcost/2+C
=-x√(1-x²)/2+arcsinx/2+C
代入①得原式=1/4*[2x²arcsinx+x√(1-x²)-arcsinx]+C