Feliciano and Uy Integral Calculus answers PDF
| Title | Feliciano and Uy Integral Calculus answers |
|---|---|
| Author | John Siegfred Alulod |
| Course | BS Mathematics |
| Institution | Pamantasan ng Lungsod ng Maynila |
| Pages | 114 |
| File Size | 3 MB |
| File Type | |
| Total Downloads | 353 |
| Total Views | 511 |
Summary
EXERCISE 9 BASIC INTEGRATION FORMULAS1. 6 π₯ 2 β 4 π₯+ 5 ππ₯= 6 π₯3 3 β4 π₯ 2 2 + 5π₯+π= πππβ πππ+ππ+π3. π₯( π₯β1)ππ₯= π₯ π₯β π₯ ππ₯= π₯3 2 ππ₯ β π₯ππ₯= ππππ πβ ππππ+π5. 2 π₯2 +4π₯β 3 π₯ 2 ππ₯= 2 + 4 πβ π 32 ππ₯= 2 ππ₯+ 4 π₯ππ₯ β π₯ 32 ππ₯= 2 π₯+ 4 ππ₯π₯ β 3 π₯β 1 β 1 ππ₯= ππ+ππππ+ ππ+π7. π₯3 β 8 π₯β 2 ππ₯ = Factor, (x-c), c = 2 P(c)...
Description
EXERCISE 9.1
BASIC INTEGRATION FORMULAS
1. 6π₯ 2 β 4π₯ + 5 ππ₯ =
β
6π₯ 3 3
7.
+ 5π₯ + π
4π₯ 2 2
3. π₯( π₯ β 1)ππ₯
=
= π₯ 2 ππ₯ β π₯ππ₯ 3
5.
ππ β π
2π₯ 2 +4π₯β3 π₯2
ππ₯
= 2+
4
= 2ππ₯ + = 2π₯ + 4
π π π π
=
+π
=
9. π
β
4
ππ₯
π₯
π₯
3 π2
ππ₯
ππ₯ β
β
= ππ + ππππ +
π
π
(π 2 +2π+4)(πβ2) (πβ2)
= (π₯ 2 + 2π₯ + 4)ππ₯
= π₯ π₯ β π₯ ππ₯ π π
ππ₯
= Factor, (x-c), c = 2 P(c) = 0 β the (x-c ) is the factor P(c) = 0 2 1 0 0 -8 2 4 8 12 4 0
= πππ β πππ + ππ + π
=
π₯ 3 β8 π₯β2
3 ππ₯ π₯2
3π₯ β1 β1
+π
ππ₯
π₯3 3
ππ π
+
2π₯ 2 2
+ 4π₯ + π
+ ππ + ππ + π
π₯ 4 β 2π₯ 3 + π₯ 2 ππ₯
= π₯ 2 ππ₯ β 2π₯
=
ππ π
β
π
ππ π
+
ππ π
2 3
ππ₯ + π₯ππ₯
+πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
1
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
1. 2 β 3π₯ ππ₯
5.
ππ’
Let u = 2 - 3x
βππ’ = ππ₯ 3
= π’ (β 1
2
ππ’ 3
ππ₯
= β3
1 1 = β π’ 2 ππ’ 3
=β
=β
3
1 2π₯ 2 3 3
+π
π π πβππ π
π
+π
3. π₯ 2 (2π₯ 3 β 1)4 ππ₯ Let u = 2π₯ 3 β 1 ππ’ = 6π₯ 2 ππ₯
ππ’ = π₯ 2 ππ₯ 6
= π₯ 2 (2π₯ 3 β 1)4 ππ₯ ππ’ = (π’4 )( ) 6
=
= = =
Let u = π₯ 2 + 3π₯ + 4 ππ’ = (2π₯ + 3)ππ₯ =
)
(2 π₯ +3)ππ₯ π₯ 2 +3π₯+4
ππ’
ππ₯
= 2π₯ + 3
ππ’
π’
= πππ’ + π
= π₯π§ ππ + ππ + π + π π₯ 2 ππ₯
(π₯ 3 β1)4
7.
Let u = π₯ 3 β 1 ππ’ = π₯ 2 ππ₯ 3
= = = =
ππ’ ππ₯
= 3π₯ 2
ππ’ 3 π₯4
1 3
π’β4
1 π’ β3 3 β3 π’ β3 β9
+π
+π
= β π(ππ βπ)π + π π
1 4 (π’ )ππ’ 6
1 π’5 +π 6 5 π’5
30
+π
(πππ β π)π +π ππ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
2
EXERCISE 9.2 ππ₯ π₯ππ 2 π₯
9.
Let u = πππ₯ ππ₯ = ππ’
ππ₯ π₯
ππ’ =
INTEGRATION BY SUBSTITUTION 13. cos 4 π₯ sin π₯ππ₯ Let u = cos π₯
1 π₯
= π’4 βππ’ =β
= π’β2 ππ’ =
π’ β1 β1
= β
π
+π
ππ₯
π π₯ β1
11.
Let u = π π₯
= ( =
=β
+π
πππ
1 π’β1
+π
ππ¨π¬π π + π
π
Let u = 3π₯
ππ’ = π π₯ ππ₯
β π’ )ππ’ 1
1 1 ππ’ β π’ π’β1
ππ£ = ππ’ π’ 1 π’
π’5 5
15. 1 + 2 sin 3π₯ πππ 3π₯ππ₯
Let v = π’ β 1
=
= βsin π₯
= - π’4 ππ’
1 ππ’ π’2
=
ππ₯
ππ’ = βsin π₯ ππ₯
1 ππ₯ ( ) ππ 2 π₯ π₯
=
ππ’
ππ£ β
ππ’
= ln (1 β π π₯ )
= 1 + 2 sin π’ πππ π’ (
;π’ =
ln|π π₯ |
ππ’
3
1 + 2 sin π’ πππ π’ππ’
1 3
)
Let v =1 + 2 sin π’
1 ππ£ π’
= ln|π’ β 1| β
ππ’ = ππ₯ 3
=
= πππ’ β πππ’ + π π’ =π’β1
ππ’ =3 ππ₯
ππ£
ππ₯
+π
= π₯π§ π β ππ β π + π
ππ’
=
= 2πππ π’ ;
ππ£ 2
= πππ π’ππ’
1 + 2 sin 3π₯ πππ 3π₯ππ₯ π£
1
ππ£
=
1 [ 3
=
1 2π£ 2 6 3
=
(π+πππππ)π π
3
2
( 2 )] +π π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
3
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
π ππ 2 π₯ππ₯ ` π+π π‘πππ₯
17.
Let u = π + π π‘πππ₯
ππ’ ππ₯
= =
=π
ππ’
π
sin π₯ πππ π₯
ππ’ π
π + πππππ + π
Let u = π‘ππ3π₯
= 3π ππ 2 3π₯ ;
= π’
=
1
[ 3
π (π₯)
= ππ₯ππ₯ +
π
π₯
π π₯
π(π₯)
* using synthetic division
-12 -8
π‘ππ3π₯ π ππ2 3π₯ππ₯
1
π (π₯ )
π₯+4
-4 3 14 13
π π₯π§ π
19.
ππ’ ππ₯
= π ππ2 π₯ππ₯
π’
1 ππ’ π’ π
=
;
=
ππ₯
3π₯ 2 +14π₯+14
21.
ππ’ 2( 3 3
2π’ 2 3
ππ’ 3
)
]+π 3
=
1 2tan 3π₯ 2 3 3
=
π(πππ§ ππ)π π
π
= π ππ 2 3π₯ππ₯
3 2 5 - R(x) π π₯ = 3π₯ + 2
π₯ + 4 = πππππππππ‘ππ π(π₯) = (3π₯ + 2) ππ₯ +
ππ₯
For the second integral : πππ‘ π’ = π₯ + 4
;
= (3π₯ + 2)ππ₯ + 5 3π₯ 2
=[ =
+π
π₯+4
5
2
πππ π
ππ¦ = 1 ; ππ’ = ππ₯ ππ₯ ππ’
π’
+ 2π₯ + 5πππ’ + π]
+ ππ + ππ₯π§(π + π) + π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
4
EXERCISE 9.2
π₯ 5 β2π₯ 3 β2π₯ π₯ 2 +1
23. π₯2
INTEGRATION BY SUBSTITUTION
ππ₯
π₯ 3 β 3π₯ +1 β 2π₯ 3 β 2π₯ π₯5 + π₯3 β3π₯ 3 β 2π₯ β3π₯ 3 β 3π₯ π₯ π₯5
π(π₯)
dx = π π₯ ππ₯ +
π(π₯)
= π₯ 3 β 3π₯ ππ₯ +
=4 β π₯4
3π₯ 2 2
π₯
π₯ ππ₯ π₯ 2 +1
+
π
(π₯)
π(π₯)
π₯ 2 +1
ππ₯
ππ₯
For the 2nd term Let u = x2+1 ππ’ = 2π₯ ππ₯
ππ’ = π₯ππ₯ 2 =4 β π₯4
=
ππ π
3π₯ 2
β
2
+
πππ π
ππ’ 2
π’
+ π π₯π§ ππ + π + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
5
EXERCISE 9.3
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
π ππ5π₯π‘ππ5π₯ππ₯
πΏππ‘ π’ = 5π₯
1.
ππ’ ππ₯
ππ’ 5
=5
7.
= ππ₯
= π πππ’π‘πππ’
ππ’
= =
5
= 5 π πππ’π‘πππ’ππ’
=
=
3.
1
5
π
π
π πππ’ + π
π ππ 2 π₯
ππ₯
π πππ₯ π ππ 2 π₯
=
1 π πππ₯
=
ππ₯ +
πππ π₯ π ππ 2 π₯
ππ’ ππ₯
ππ₯
ππ₯
; Let u= 2 π₯
ππ’
ππ₯
=
= =2 =2 =2
1
1
2ππ’ = ππ₯
1 2
π πππ’πππ‘π’ 2ππ’
ππ’
π πππ’ πππ‘π’ ππ’
cos 3 π₯ 1+π πππ₯ ππ₯ cos 3 π₯
= πππ π₯ ; ππ’ = πππ π₯ππ₯
= π πππ₯ +
= β ππ ππππ + ππππ β ππππ + π
1
1βsin 2 π₯
= π πππ₯ + π’ππ’
ππ₯ + πππ‘π₯ππ ππ₯ππ₯
sin π₯ cot π₯ 2 2
(co s 3 π₯+cos 3 π₯π πππ₯ )ππ₯
πΏππ‘ π’ = π πππ₯
= ππππ +
= ππ πππ₯ + πππ‘π₯ππ ππ₯ππ₯
5.
(1βπ πππ₯ )(1+π πππ₯ )
= π πππ₯ + πππ π₯π πππ₯ππ₯
πππππ + π
π πππ₯ +πππ π₯
1+π πππ₯ . 1+π πππ₯ 1βπ πππ₯ (cos 3 π₯ ) 1+π πππ₯ ππ₯
= πππ π₯ 1 + π πππ₯ ππ₯
1
=
cos 3 π₯ππ₯
9.
π’2 2
+π
π¬π’π§π π π
+π
1 + π‘πππ₯ 2 ππ₯
= (1 + 2π‘πππ₯ + tan2 π₯)ππ₯
= [2π‘πππ₯ + (1 + tan2 π₯)]ππ₯
= 2 π‘πππ₯ππ₯ + sec 2 π₯ππ₯
= 2 βππ|πππ π₯| + π‘πππ₯ + π
= βπ ππ |ππππ| + ππππ + π
πππ π’ ) π πππ’
π πππ’ (
1 πππ π’
(ππ’)
= 2 π πππ’ππ’
= πππ ππππ + ππππ + π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
6
EXERCISE 9.3
11.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
πππ 6 π₯ππ₯ cos 2 3π₯
15.
Let u = 3x ; 2u = 6x
ππ’
ππ₯
=
=3 ; πππ 2π’
2
=3
ππ’
cos 2 π’ πππ π’
1
πππ π’ πππ π’
= 3 π πππ’ππ’ 2
=
3
=
π ππ 2 π₯ππ₯
=
1 πππ π₯
ππ₯
= π πππ₯ππ₯
= ππ ππππ + ππππ + π
1
.
ππ’ 2
ππ’ 2
= ππ₯
2
π‘πππ’ππ’
ππ₯
2π πππ₯πππ π₯ππ₯
ππ₯ πππ π₯
ππ₯
ππ₯
π = β π₯π§ πππππ + π π
2π πππ₯ππ π 2 π₯
=
=2
π πππ’ πππ π’
π
(2π πππ₯πππ π₯ )πππ π₯
ππ₯
πππ 2π₯
ππ’ ππ₯
π
=
2π πππ₯πππ π₯ πππ 2π₯
Let u = 2x
= π₯π§ πππππ + πππππ + π
13.
ππ₯
π ππ 2 π₯
=
ππ’
π ππ 2 π₯πππ 2π₯
(4π πππ₯πππ π₯ )(π πππ₯πππ π₯ ) 2π πππ₯πππ π₯ πππ 2π₯
=
= ππ₯
ππ’ 3
4 sin 2 π₯ππ π 2 π₯
π ππ 3 π₯π‘ππ 3π₯
17.
Let u = 3x
ππ’ ππ₯
=
=3 ππ’ 3
ππ’ 3
= ππ₯
π πππ’π‘πππ’
=3 ππ ππ’ + π 1
= β π πππππ + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
7
EXERCISE 9.4 ππ₯ π 2π₯
1.
π β2π₯ dx
=
INTEGRATION OF EXPONENTIAL FUNCTIONS
ππ’
ππ’ πππ‘ π’ = 2π₯ ; = β2 ; β 2 = ππ₯ ππ₯ ππ’ = π π’ (β ) 2 =β
π π’ ππ’
1
=β π π’ + π 2 2
1
=β
2π π’ 1
+π
=β π (πβππ) + π π
3. π π ππ4π₯ πππ 4π₯ππ₯ ππ’ π π πππ’ πππ π’(
1 4
=
4
π π πππ’ πππ π’ππ’
=
= =
4 1 4
π π£ ππ£
π +π π£
ππππππ π
+π
=
3π₯
2ππ’ ππ’ ( 3
;
2
)
2
ππ₯
2ππ’ = ππ₯ 3
)
= 3 π π’ ππ’ 2
π
+π
3π₯ 2
=
2 3
=
π πππ π
+π
πππ‘ π’ = 3 β 2π₯ ;
ππ£ = cos π’ ; ππ£ = πππ π’ππ’ πππ‘ π£ = π πππ’ ; ππ’ 1
πππ‘ π’ =
3π₯
7. 53β2π₯ ππ₯
ππ’ ππ’ = ππ₯ πππ‘ π’ = 4π₯ ; = 4; 4 ππ₯
=
π 3π₯ ππ₯ = π
5.
= 5π’ (β =β
1 2
= β2 =β
ππ’
2
5π’ ππ’
1 5 3β2π₯ ππ5
ππβππ ππππ
)
ππ’ ππ’ = β2 ; β = ππ₯ ππ₯ 2
+π
+π
9. 3π₯ 2π₯ ππ₯
π π₯ π π₯ = (ππ)π₯ = 6π₯ ππ₯ =
ππ
πππ
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
8
EXERCISE 9.5
INTEGRATION OF HYPERBOLIC FUNCTIONS
π πππ 3π₯ β 1 ππ₯
1.
Let u = 3π₯ β 1 = 3 ; ππ₯ =
ππ’ ππ₯
= π πππ π’ (
= =
ππ’
3.
π πππ π’ππ’
1 3
ππ’
3
π ππππ π
=
)
= ππππ (πππ) + π
ππ β π + π
ππ ππ2 1 β π₯ 2 π₯ππ₯ Let u=1 β π₯ 2
β
ππ’ ππ₯
= -2π₯
ππ’ = π₯ππ₯ 2
= ππ ππ2 π’(β
1 ππ’ ππ₯ πππ‘ π’ = πππ₯ ; ππ₯ = ; ππ’ = π₯ π₯ π πππ2 π’ππ’
3
πππ π π’ππ’ + c
1
=
3
π ππ π 2 πππ₯ ππ₯ π₯
5.
ππ’
2
=β 2 ππ ππ2 π’ππ’
= π‘ππππ’ + π
7.
ππ ππ
1 π₯ 2
Let u = 2 π₯ ; 1
πππ‘π 2 π₯ππ₯
ππ’
ππ₯
1
=
1 2
= 2 ππ πππ’ πππ‘ππ’ππ’
; 2ππ’ = ππ₯
= 2(βππ πππ’ + π)
= βπππππ π π + π π
)
1
=β 2 (βπππ‘ππ’ + π) 1
= π ππππ π β ππ + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
9
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
1. Given slope 3π₯ 2 + 4 ππ¦ = 3π₯ 2 + 4 ππ₯
ππ¦ = 3π₯ 2 + 4 ππ₯
ππ¦ = 3π₯ 2 + 4 ππ₯
π¦=
3π₯ 3 3
+ 4π₯ + π
π = πππ + ππ + π
3. Given slope ππ¦ π₯ + 1 = ππ₯ π¦ β 1
π₯+1 π¦β1
π¦ β 1 ππ¦ = π₯ + 1 ππ₯
π¦ 2 β 2π¦ =
π₯2
+π₯+π2
π¦ 2 β 2π¦ = π₯ 2 + 2π₯ + 2π 2
ππ β ππ + ππ + ππ + ππ = π 5. Given slope ππ¦ 1 = ππ₯ π₯π¦ π¦ππ¦ =
π¦2 2
=
ππ₯
ln π₯ 2 2
π₯π¦ 1
7. Given slope
ππ¦
π¦2 = π₯
ππ₯ ππ¦ ππ₯ = 2 π₯ π¦ β
+π 2
ππ = ππππ + ππ
π₯
, through 1,4
1 = πππ₯ + π 4
β ln π₯ β
1 =π 4
1 β ln 1 β = π 4
π=β
1 4
β ln π₯ β
1 1 + = 0 4π¦ π¦ 4
β4π¦ ln π₯ β 4 + π¦ = 0 ππ π₯π§ π β π + π = π
9. Given slope π¦, through 1,1 ππ¦ = π¦ ππ₯
π¦ β 2 ππ¦ = ππ₯ 1
π¦2
=π₯+π
1
π₯
π¦2
1 2
2π¦
1
2=
π₯+π
When π₯ = 1 , π¦ = 1
21 =1+π ; π =1 2π¦
1
2
=π₯+π
2
ππ = π + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
10
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
11. Given slope π₯ β2 , through 1,2 1 ππ¦ = 2 ππ₯ π₯ ππ¦ =
ππ₯
π₯2
1 π¦ =β +π π₯ 1 2=β +π 1 2 = β1 + π π=3
π¦=β
1 π₯
+3x
π₯π¦ = β1 + 3π₯
ππ β ππ + π = π
v = -32t + vo when t = 1 sec, s=h=48ft h=-16t2+ vot + c1 48 = -16(1)2 + vo(1) + c2 64 - vo = c2 When t = 0, s = 0, c2 = 0 s = -16t2 + vot when t = 1 sec, s = 48 s = -16t2 + c1t 48 = -16(1)2 + c1(1) c1=64 s=-16t2 + 64t v = -32t + 64 @ max, v = 0
13. 0 = -32t + 64 a=-32 ft/sec2 32t=64 a=-2
ππ¦ = β32 ππ‘
ππ£ = β 32ππ‘
t = 2 sec s = -16t2 + 64t s = -16(2)2 + 64(2) s = 64ft
v=-32t+c
ππ = β32π‘ + π1 ππ‘
ππ = (β32π‘ + π1 )ππ‘
s=16t2 + c1t + c2 when t = 0, v = vo v=-32t + c1 vo= -32(0) + c1 vo =c1 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
11
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
15. a = 32ft/sec2 a = 32
ππ£ = 32 ππ‘
ππ£ = 32ππ‘
v = 32t + c1
ππ = 32π‘ + π1 ππ‘
ππ = 32π‘ + π1 ππ‘
S = 16t2 + c1 + c2
when t = 0, v = 0 c1 = 0 v = 32t when t = 0 , s = 0 c2 = 0 s = 16t2 π‘= π‘=
400 16 20 4
t = 5 sec v = vt *since it is a free falling body, its velocity is ( - ) vt = -32t vt = -32(5) vt = -160 ft/sec
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
12
EXERCISE 10.1
PRODUCT OF SINES AND COSINES
1. Κ sin 5π₯ sin π₯ ππ₯
5. Κ cos 3π₯ β 2π cos π₯ + π ππ₯
= [cos π’ β π₯ β cos(π’ + π£)]ππ₯
πππ‘ π’ = 3π₯ β 2π
= 2 sin π’ sin π£ ππ₯ π’ = 5π₯
1 = Κ[cos π’ + π£ + cos (π’ β π£ )]ππ₯ 2
π£ =π₯+π
π£=π₯
1 = Κ[cos 5π₯ β π₯ β cos (5π₯ + π₯)]ππ₯ 2
ο·
1 = Κ[cos 4π₯ β cos 6π₯]ππ₯ 2
ο·
1 = [Κ cos 4π₯ππ₯ β Κ cos 6π₯ππ₯ 2
= =
1 1 [ sin 4π₯ 2 4
1 β 6 sin 6π₯
πππ ππ πππππ β +πͺ π ππ
]+πΆ
=β
1 5
= 2π₯ β 3π
= βπππ 4π₯
πππ cos 2π₯ β 3π = cos 2π₯πππ 3π + sin 2π₯π ππ 3π = β cos 2π₯
= 2 Κ(cos 4π₯ β cos 2π₯)ππ₯ 1
1 = Κ[sin 5π₯ + 2 + sin(3π₯ β 8)]ππ₯ 2
1
π’ β π£ = 3π₯ β 2π β π₯ + π
πππ cos 4π₯ β π = cos 4π₯πππ π + π ππ4π₯π πππ
1 = Κ [sin 9x β 3 + x + 5 + sin 9π₯ β 3 β π₯ β 5 ππ₯ 2
= [β cos π§ 2
= 4π₯ β π
1 = Κ[cos 4π₯ β π + cos(2π₯ β 3π)]ππ₯ 2
3. Κ sin 9π₯ β 3 cos π₯ + 5 ππ₯
πππ‘ π§ = 5π₯ + 2 ππ§ =5 ππ₯ ππ§ = ππ₯ 5
π’ + π£ = 3π₯ β 2π + π₯ + π
; πππ‘ π€ = 3π₯ β 8 ππ€ =3 ; ππ₯ ππ€ ; = ππ₯ 3
= 2 [β 4 sin 4π₯ β 2 sin 2π₯] + πΆ 1
1
1
π π = β π¬π’π§ ππ β π¬π’π§ ππ + πͺ π π
β 3 πππ π€] + πΆ 1
π π πππ ππ β π + πͺ πππ ππ + π β ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
13
EXERCISE 10.1
PRODUCT OF SINES AND COSINES
7. 4 π ππ 8π₯ πππ 3π₯ππ₯
= 2Κ[sin 8π₯ + 3π₯ + π πππ₯ 8π₯ β 3π₯ ππ₯
9. 5 π ππ 4π₯ + 3 π ππ 2π₯ β 6 ππ₯
5 (π’ + π£)]ππ₯ = Κ[cos π’ β π£ β cos 2 π πππ‘ π’ = 4π₯ + ; 3
= 2Κ[π ππ11π₯ + sin 5π₯]ππ₯
πππ‘ π’ = 11π₯ ; πππ‘ π£ = 5π₯ ππ’ = 11 ; ππ₯
ππ’ = ππ₯ ; 11
ππ£ =5 ππ₯
ππ£ = ππ₯ 5
= 2[β 11 cos 11π₯ β 5 cos 5π₯ ] + πΆ 1
1
π π = β ππ¨π¬ πππ β πππππ + πͺ π ππ
π
π
ο·
π’ β π£ = 4π₯ +
π 3
β 2π₯ β
π 6
ο·
π’ + π£ = 4π₯ +
π 3
+ 2π₯ β
π 6
ο·
= 2π₯ + π/2 π π£ = 2π₯ β 6 = 6π₯ + π/6
π π 5 β πππ 6π₯ + = Κ[πππ 2π₯ + ]ππ₯ 2 2 6
πππ cos 2π₯ + 2 π
= βπ ππ2π₯
= cos 2π₯πππ
ο·
πππ cos 6π₯ + 6 π
π π β sin 2π₯ sin 2 2
π π β π ππ 6π₯ π ππ 6 6 1 3 πππ 6π₯ β π ππ 6π₯ = 2 2 = πππ 6π₯πππ
=
3 5 1 Κ[β π ππ 2π₯ β πππ 6π₯ + π ππ 6π₯ ]ππ₯ 2 2 2
5 = 2 [2 πππ 2π₯ β
=
1
3
πππ ππ β
12
π
π
π ππ 6π₯ β
ππ ππ
1 12
π ππ₯ 6π₯ + πΆ
πππ ππ β ππ πππ ππ + πͺ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
14
EXERCISE 10.2
POWER OF SINES AND COSINES
1. π ππ3 π₯πππ 4 π₯ππ₯; ππ¦ πΆππ π πΌ = π ππ4 π₯πππ 4 π₯π πππ₯ππ₯
= (1 β πππ 2 π₯)2πππ 4 π₯π πππ₯ππ₯
= (1 β 2πππ 2 π₯ + πππ 4 π₯)πππ 4 π₯π πππ₯ππ₯
= (πππ 4 π₯ β 2πππ 6 π₯ + πππ 8 π₯)π πππ₯ππ₯ Let u = cosx
-ππ’ = π πππ₯ππ₯
= - (π’4 β 2π’...
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