integration

d/dx[tan(x/2)]
=(1/2)[sec(x/2)}^2
=1/{2[cos(x/2)]^2}
=2/[2(1+cosx)]
=1/(1+cosx)
=RHS

d/dx ( x + sinx )
= 1 + cosx

............

d/dx(uv) = v(du/dx) + u(dv/dx)
v(du/dx) = d/dx(uv) - u(dv/dx)

Therefore ,

積分符號[ v(du/dx) ] dx = 積分符號 d/dx(uv) dx - 積分符號 u(dv/dx) dx
積分符號[ v(du/dx) ] dx = uv - 積分符號 u(dv/dx) dx

Now ,
putting v in (x+sinx) and du/dx = 1/(1+cosx) => u = tan(x/2)


積分符號[(x+sinx)/(1+cosx)]dx
= tan(x/2)(x+sinx) - 積分符號 u(dv/dx) dx
= tan(x/2)(x+sinx) - 積分符號 tan(x/2)[1 + cosx ] dx

.........

tan(x/2)[1 + cosx ]
=[(1+cosx)*sqtr(1-cosx)]/sqrt(1+cosx)
=sqrt(1+cosx)*sqtr(1-cosx)]
=sqrt[1-(cosx)^2]
=sqrt(sinx)^2
=sinx

tan(x/2)(x+sinx) - 積分符號 tan(x/2)[1 + cosx ] dx
= tan(x/2)(x+sinx) - 積分符號 sinx dx
= tan(x/2)(x+sinx) + cosx + C

2008-07-02 18:43:09 補充：
Let v = x + sinx , dv/dx = 1+ cosx

2008-07-03 09:53:22 補充：
多謝下面詳細地解析我是怎樣d tan(x/2)

不過我覺得考生應該要記住
(d/dx) :

(d/dx) sinx = cosx
(d/dx) cosx = -sinx
(d/dx) tanx = (secx)
(d/dx) secx = tanxsecx
(d/dx) cscx = -cotx cscx
(d/dx) cotx = -cscx

咁樣考試時就不會用咁多時間計呢到 ~~

我諗書是不會印晒出來,要我們找出來

a) Same as 001 and 002, so not to repeat here.
b) S(x+sinx)dx/(1+ cosx) = Sxdx/(1+cosx) + Ssinxdx/(1+cosx) = A + B.
For part B, since d(1+cosx)= sinxdx, therefore,
Ssinxdx/(1+cosx) = Sd(1+cosx)/(1+cosx) = ln(1+cosx) +C.
For part A, using integration by parts. u=x and v=tan(x/2).
Sxd[tan(x/2)] = xtan(x/2) - Stan(x/2)dx.....................(1)
Using result in (a) Sxd[tan(x/2)] = Sxdx/(1+cosx) ......(2)
For Stan(x/2)dx= S[sin(x/2)/cos(x/2)]dx, again by method of substitution of d[cos(x/2)] = (1/2)sin(x/2)dx,
S[sin(x/2)/cos(x/2)]dx = S2d[cos(x/2)]/cos(x/2) = 2ln[cos(x/2)] + C.
Put this result into (1) and (2), we get
Sxdx/(1+cosx) = xtan(x/2) - 2ln[cos(x/2)] + C.
So A + B = xtan(x/2) - 2ln[cos(x/2)] + ln(1+cosx) + C.
= xtan(x/2) + ln[(1+cosx)/cos^2(x/2)] + C.
= xtan(x/2) + ln[2cos^2(x/2)/cos^2(x/2)] + C
=xtan(x/2) + ln2 + C.
=xtan(x/2) + C.

2008-07-03 15:45:09 補充：
Correction: d(1+cosx) should be -sinxdx. And d[cos(x/2)] should be (-1/2)sin(x/2)dx. But these does not affect the final answer.

比上面做慢左好多,但a part不一樣,所以給你參考`


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