求达人解数学(定积分)的题!

问题描述:

求达人解数学(定积分)的题!
鞠躬~
1个回答 分类:数学 2014-09-28

问题解答:

我来补答
很有礼貌
2.
tan(x/2)
=sin(x/2)/cos(x/2)
=(1-cosx)^(1/2)/(1+cosx)^(1/2)
={(1-cosx)^(1/2)/(1+cosx)^(1/2)}*
{(1+cosx)^(1/2)/(1+cosx)^(1/2)}
=sinx/(1+cosx)
1+tan(x/2)=(1+sinx+cosx)/(1+cosx)
∫(0,pi/2) dx/(1+sinx+cosx)
= ∫(0,pi/2) (1/2)dx/{(1/2)(1+cosx)(1+tan(x/2))}
= ∫(0,pi/2) [{sec(x/2)}^2d(x/2)]/[1+tan(x/2)]
= ∫(0,pi/2) d[tan(x/2]/(1+tan(x/2)]
=ln (1+tan(x/2)) (x=0,pi/2)
3.
∫x^2 cosx dx =x^2sinx - ∫2xsinxdx (integration by parts)
= x^2sinx +2xcosx- ∫2cosxdx
4.
f= ∫ (0,1) dx/(1-x^2)^(1/2)
siny=x
cosydy=dx
f= ∫ (0,pi/2) cosy dy/(cosy)=y=arsin(x) [x=0,1]
1.
f=∫(0,ln5){e^x (e^x-1)^(1/2))/(e^x+3) dx
y=e^x
dy=e^x dx
f= ∫(1,5) {y-1)^(1/2)/ (y+3) dy
=(1/(2-3))[2(x-1)^(1/2) -4 ∫dx/[(x+3)(x-1)^(1/2) [x=1,5]
=-{2(x-1)^(1/2)-4arctan[(x-1)/2]} [x=1,5)
 
 
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