Find the integrals of
1/[ x^(1/2) + x^(1/3) ]
Square root of { sqrt(2) + sin[sqrt(x)] + cos[sqrt(x)]}
再來兩個不定積分
2008-08-02 3:31 am
回答 (4)
2008-08-06 11:57 am
✔ 最佳答案
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參考資料：
修改自http://i117.photobucket.com/albums/o61/billy_hywung/Aug08/Crazyint1.jpg及http://i187.photobucket.com/albums/x22/cshung/7008080102888.png + my maths knowledge
2008-08-02 10:27 pm
Q2 : 經開方後出現絕對值
sqrt[ 1 + sqrt(丌/4 + u)]
= | sin(丌/8 + y/2) + cos(丌/8 + y/2) |
2008-08-02 14:30:37 補充：
係
sqrt[ 1 + sin(丌/4 + y)]先岩
= sqrt[ sin^2 u + cos^2 u + 2sin(u)cos(u) ]
= sqrt( sinu + cosu )^2
= | sinu + cosu |
u係咩任人定
2008-08-03 10:44:19 補充：
那咩時候係咩case?我唔識這樣....
有沒有什麼不等式
sqrt[ 1 + sqrt(丌/4 + u)]
= | sin(丌/8 + y/2) + cos(丌/8 + y/2) |
2008-08-02 14:30:37 補充：
係
sqrt[ 1 + sin(丌/4 + y)]先岩
= sqrt[ sin^2 u + cos^2 u + 2sin(u)cos(u) ]
= sqrt( sinu + cosu )^2
= | sinu + cosu |
u係咩任人定
2008-08-03 10:44:19 補充：
那咩時候係咩case?我唔識這樣....
有沒有什麼不等式
2008-08-02 8:22 pm
http://i187.photobucket.com/albums/x22/cshung/7008080102888.png
(For Q2)
There is a bug in the proof, can you find it? :p
2008-08-02 18:48:37 補充：
既然你搵到了咁即係你識做啦
分番晒case就得,太煩唔做了。
2008-08-05 20:12:39 補充：
Let S = sin(y/2 + pi/8)
C = cos(y/2 + pi/8)
Case 1: S is positive, C is positive, S >= C => |S - C| = S - C
Case 2: S is positive, C is positive, S < C => |S - C| = C - S
Case 3: S is positive, C is negative => |S - C| = S + C
Case 4: S is negative, C is positive => |S - C| = -S - C
2008-08-05 20:12:46 補充：
Case 5: S is negative, C is negative, S >= C => |S - C| = S - C
Case 6: S is negative, C is negative, S < C => |S - C| = C - S
Notice, Case 1 = Case 5, Case 2 = Case 6, there are really just 4 cases.
You can represent the solution with all these cases.
(For Q2)
There is a bug in the proof, can you find it? :p
2008-08-02 18:48:37 補充：
既然你搵到了咁即係你識做啦
分番晒case就得,太煩唔做了。
2008-08-05 20:12:39 補充：
Let S = sin(y/2 + pi/8)
C = cos(y/2 + pi/8)
Case 1: S is positive, C is positive, S >= C => |S - C| = S - C
Case 2: S is positive, C is positive, S < C => |S - C| = C - S
Case 3: S is positive, C is negative => |S - C| = S + C
Case 4: S is negative, C is positive => |S - C| = -S - C
2008-08-05 20:12:46 補充：
Case 5: S is negative, C is negative, S >= C => |S - C| = S - C
Case 6: S is negative, C is negative, S < C => |S - C| = C - S
Notice, Case 1 = Case 5, Case 2 = Case 6, there are really just 4 cases.
You can represent the solution with all these cases.
2008-08-02 6:32 pm
收錄日期: 2021-04-23 18:02:02
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