→ ElvinN:推 04/16 21:33
※ 引述《newlife148 (小夏)》之銘言：
: 小弟的期中考題目..
: 我貼在相簿裡面有四題..懇請各位幫忙解惑
: http://www.wretch.cc/album/show.php?i=uniquepledge&b=2&f=1262165804&p=1
: http://www.wretch.cc/album/show.php?i=uniquepledge&b=2&f=1262165803&p=0
: 不會縮網址..抱歉
dx
----------
1.令√(x+2) = u du = 2√(x+2)
dx= 2udu
√(x+2) u u u u u
∫10 dx = ∫10 2udu = 2u 10 -∫2 10 du = 2u 10 - 2 10
------ ----- ------ --------
ln10 ln10 ln10 (ln10)^2
√(x+2) √(x+2)
=2√(x+2) 10 2√(x+2) 10
------------------ - ------------------ + C
ln10 (ln10)^2
3 2
2.令u=x^(1/3) u =x , 3u du = dx
2 2
∫sin(x^(1/3))dx = ∫sin(u)3u du = -3u cosu +∫6ucosudu
2 2
=-3u cosu -6usinu+∫6sinudu = -3u cosu -6usinu -6cosu
=-3x^(2/3)cos(x^(1/3)) -6x^(1/3) sin(x^(1/3))-6cos(x^(1/3))+C
3.令z=tan(x/2)
sinx=2z/(1+z^2) cosx=(1-z^2)/(1+z^2)
(1+z^2)dz = dx
∫1/(2sinx-cosx+1) dx = ∫(1+z^2)/(4z/(1+z^2) -(1-z^2)/(1+z^2) +1) dz
= ∫(1+z^2)^2/(2z^2+4z)dz = ∫1/2 z^2 -z+3 +(1-6z)/(2z^2+4z)dz
=1/6 z^3 -1/2 z^2 +3z +∫1/4z -13/4(z+2)dz
=1/6 z^3 -1/2 z^2 +3z + 1/4 ln(z) -13/4 ln(z+2)
3 2
=1/6tan (x/2)-1/2 tan (x/2)+3tan(x/2) +1/4 ln(tan(x/2)) -13/4 ln(2+tan(x/2))+C
4.∫1/(2cosx-sinx+1) dx = ∫(1+z^2)/(2(1-z^2)/(1+z^2) -2z/(1+z^2) +1) dz
=∫(1+z^2)^2/(3-z^2-2z)dz = ∫-z^2+2z-9 +(-12z+28)/(3-z^2-2z)dz
=∫-z^2+2z-9-16/(z+3)+4/(z-1)dz
=-1/3 z^3 + z^2 -9z -16ln(z+3) +4ln(z-1)
3 2
= -1/3 tan (x/2) + tan (x/2) -9tan(x/2) -16ln(3+tan(x/2)) +4ln(tan(x/2-1))+C
有錯麻煩指教
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