Download L29_8.3 Powers and products of Trigonometric functions(三角函數的

L29 8.3 Powers and products of Trigonometric functions (三角函數的次數和乘積)
Part I: For Sinx and Cosx
Part II: Other trigonometric functions
∫e
ax
cos(bx)dx = ? 類似 ∫ e x cos( x)dx = ?
By the way~定義數學的約定，定義是數學的起家，規定、準則。
§ 8.3 Powers and products of Trigonometric functions
PartⅠ: For sinx and cox
Goal:
∫ sin x
m
cos n dx = ?
caseⅠ: m or n is odd, say m=2k+1.
Q:m 跟 n 什麼樣的狀況 substitution 用的上？
A:m 跟 n 有一個是單數。
∫ sin x
2k +1
n
cos
=
dx
∫ sin x(1 − cos
2
x) k cos n dx cosx 的多項式
Let u=cosx, then du=-sinxdx
-∫(u 的多項式)du=積得出來
caseⅡ: m and n are even.
Using the following identities to reduce the power
1
2
sinxcosx= sin x cos x , sin2x=
caseⅢ: ∫ sin(mx) cos(nx)dx ,
1+ cos(2 x)
1 − cos(2 x)
, cos2x=
2
2
∫ sin(mx) sin(nx)dx , ∫ cos(mx) cos(nx)dx .(m≠n)
sinAcosB=1/2[sin(A+B)+sin(A-B)]
sinAsinB=1/2[cos(A-B)-cos(A+B)]
cosAcosB=1/2[cos(A+B)+cos(A-B)]
L29 8.3 Powers and products of Trigonometric functions (三角函數的次數和乘積)
Part I: For Sinx and Cosx
Part II: Other trigonometric functions
eg.
∫ cos
1
○
2
dx =
1 + cos(2 x) x 1
=
=
+ sin(2 x) + C
2
2 4
∫ sin
2
○
2
x cos5 dx =
= ∫ sin 2 x(1 − sin 2 ) 2 cos xdx = ∫ (sin 2 x − 2sin 4 x + sin 6 x) cos xdx =
∫ sin
3
○
2
1 3
2
1
sin x − sin 5 x + sin 7 x + C
3
5
7
x cos 2 dx =
1
1 1 − cos(4 x)
1
1
2
=
dx =
x − sin(4 x) + C
∫ [ 2 sin(2 x)] dx =
∫
4
2
8
32
∫ sin
4
○
5
xdx =
= sin x(1 − cos 2 ) 2 dx
∫ sin
5
○
4
x cos 2 xdx =
1 − cos(2 x) 1
1
2
2
2
=
∫ 2 ⋅ [ 2 sin(2 x)] dx =8 ∫ sin (2 x) − cos(2 x) sin (2 x)dx
=
1 1 − cos(4 x)
1 1 1
1
1
1
dx − ⋅ ⋅ sin 3 (2 x=
)
x − sin(4 x) − sin 3 (2 x) + C
∫
8
2
8 2 3
16
64
48
∫ sin(5x) cos(4x)dx =
6
○
1
cos(9 x) cos x
sin(9x) + sin( x)dx =
−
−
+C
∫
2
18
2
Remark:
∫ sin
n
∫ sin
n
xdx ,
∫ cos
n
xdx
2
n −1
n −1
n−2
xdx =
∫ sin ⋅ sin xdx =−sin cos x + (n − 1)∫ cos x ⋅ sin dx
=
− sin n −1 cos x + (n − 1) ∫ sin n − 2 dx − (n − 1) ∫ sin n dx
⇒
−1 n −1
(n − 1)
sin cos x +
sin n − 2 dx apply this equation for n-2.
∫
n
n
Ex:P415(2.8.16.27.29.33.34)
L29 8.3 Powers and products of Trigonometric functions (三角函數的次數和乘積)
Part I: For Sinx and Cosx
Part II: Other trigonometric functions
Part:Other trigometric functions
∫ tan
(Ⅰ)
n
xdx , ( ∫ cot n xdx )
Q:能不能製造出 sec2x？A:抽出 tan2x 轉換(sec2x-1)
tan dx ∫ tan
∫=
n
n−2
∫ tan
(sec 2=
x − 1)dx
n−2
sec 2 xdx − ∫ tan n − 2 xdx
重複一直作到次數為 1 或 0
eg.
∫ tan
6
dx
tan 5 x
= ∫ tan x(sec x − 1)dx=
− ∫ sec 2 x(sec 2 x − 1) xdx
5
1
1
1
1
= tan 5 x − tan 3 x + ∫ (sec 2 x − 1) x = tan 5 x − tan 3 x + tan x − x + C
5
3
5
3
4
2
∫ sec
(Ⅱ)
n
xdx , ( ∫ csc n xdx )
Q:能不能製造出 secxtanx？A:不能，只能製造出 sec2xtan2x (substitution 不能用)
If n is even, say n=2k
∫ sec
2k
k −1
k −1
2
2
2
xdx =
∫ sec x(tan x + 1) dx =∫ (u + 1) du
If n is odd, say n=2k+1
2k-1
2
Q:要拆成怎樣？A:sec x 和 sec x
xdx ∫ sec
∫ sec =
2k +1
2k −1
2
x sec
=
dx sec 2k −1 x tan x − ∫ (2k − 1) sec 2 k − 2 x sec x tan x ⋅ tan xdx
= sec 2k −1 x tan x − (2k − 1) ∫ sec 2 k +1 xdx + (2k − 1) ∫ sec 2 k −1 xdx
⇒ ∫ sec 2 k +1 xdx
=
1
2k − 1
tan x sec 2 k −1 x +
sec 2 k −1 dx
∫
2k
2k