Question

几道求不定积分大学数学题

求下列几个的不定积分，要过程哈 1.1/(sinx^2*cosx^2) 2.(lnx)/(x^2) 3.1/[ (x+1)*√x] 4.(1+cosx)/(x+sinx) 5.x*cos2x 6.x*e^x 7.(1-x^2)^(3/2) 8.√x/√x-1

Answer 1

1.∫1/(sinx^2*cosx^2)dx
=∫([(sinx)^2+(cosx)^2]/[(sinx)^2*(cosx)^2]dx
=∫[1/(cosx)^2+1/(sinx)^2]dx
=tanx-cotx+C.
2.∫(lnx)/(x^2)dx=-∫(lnx)d(1/x)
=-(lnx)/x+∫(1/x^2)dx
=-(lnx)/x-1/x+C.
3.∫1/[(x+1)*√x]dx=2∫1/(x+1)d(√x)
=2arctan√x+C.
4.∫(1+cosx)/(x+sinx)dx
=∫1/(x+sinx)d(x+sinx)
=ln|x+sinx|+C.
5.∫xcos2xdx=(1/2)∫xd(sin2x)
=(1/2)xsin2x-(1/2)∫sin2xdx
=(1/2)xsin2x+(1/4)cos2x+C.
6.∫x*e^xdx=∫xd(e^x)
=x*e^x-∫e^xdx=(x-1)e^x+C.
7.先做一道I=∫(1-x^2)^(1/2)dx
=x*(1-x^2)^(1/2)+∫x^2*(1-x^2)^(-1/2)dx
=x*(1-x^2)^(1/2)+∫[1-(1-x^2)]*(1-x^2)^(-1/2)dx
=x*(1-x^2)^(1/2)+arcsinx-I,
所以
I=[x*(1-x^2)^(1/2)+arcsinx]/2+C.
再计算J=∫(1-x^2)^(3/2)dx
=x*(1-x^2)^(3/2)+3∫x^2*(1-x^2)^(1/2)dx
=x*(1-x^2)^(3/2)+3∫[1-(1-x^2)]*(1-x^2)^(1/2)dx
=x*(1-x^2)^(3/2)+3I-3J.
J=[2*x*(1-x^2)^(3/2)+3*x*(1-x^2)^(1/2)+3*arcsinx]/4+C.
8.本题关键一是是使用常用公式∫1/√(1+u^2)du=ln[u+√(1+u^2)]+C.
先计算I=∫√(1+u^2)du
=u*√(1+u^2)-∫u^2/√(1+u^2)du
=u*√(1+u^2)-∫[(u^2+1)-1]/√(1+u^2)du
=u*√(1+u^2)-∫√(1+u^2)du+∫1/√(1+u^2)du
=u*√(1+u^2)-I+ln[u+√(1+u^2)].
所以I=(1/2)*u*√(1+u^2)+(1/2)*ln[u+√(1+u^2)]+C.
J=∫√x/√(x-1)dx，令u=√(x-1)；
J=2∫√(1+u^2)du=
=u*√(1+u^2)+ln[u+√(1+u^2)]+C
=√(x-1)*√x+ln[√(x-1)+√x]+C.