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calculus book of feliciano uy solutions and answer
Typology: Schemes and Mind Maps
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3 3 −^
4 𝑥 2 2 + 5𝑥^ +^ 𝑐 = 𝟐𝒙𝟑^ − 𝟐𝒙𝟐^ + 𝟓𝒙 + 𝒄
3 (^2) 𝑑𝑥 − 𝑥𝑑𝑥
= 𝟐𝟓 𝒙
𝟓 𝟐 (^) − 𝟏𝟐 𝒙𝟐^ + 𝒄
(^2) +4𝑥− 3 𝑥 2 𝑑𝑥
= 2 + (^4) 𝑋 − (^) 𝑋^32 𝑑𝑥
= 2 𝑑𝑥 + (^4) 𝑥 𝑑𝑥 − (^) 𝑥^3 2 𝑑𝑥
− 1 − 1 𝑑𝑥
= 𝟐𝒙 + 𝟒𝒍𝒏𝒙 + 𝟑𝒙 + 𝒄
(^3) − 8 𝑥− 2 𝑑𝑥 = Factor, (x-c), c = 2 P(c) = 0 – the (x-c ) is the factor P(c) = 0 2 1 0 0 - 2 4 8 1 2 4 0
= (𝑋
(^2) +2𝑋+4)(𝑋−2) (𝑋−2)
= (𝑥^2 + 2𝑥 + 4)𝑑𝑥
= 𝑥^
3 3 +^
2 𝑥 2
2
𝟑 𝟑 +^ 𝒙
2 (^3) 𝑑𝑥 + 𝑥𝑑𝑥
𝟑 𝟑 −^
𝟔 𝟓𝟑 𝟓 +^
𝒙𝟐 𝟐 +^ 𝑪
Let u = 2 - 3x𝑑𝑢𝑑𝑥 = − 3
−𝑑𝑢 3
1 (^2) (− 𝑑𝑢 3 )
1 (^2) 𝑑𝑢
3 2 3 +^ 𝑐
𝟑 𝟐 𝟗 +^ 𝒄
Let u = 2𝑥^3 − 1
𝑑𝑢 𝑑𝑥 = 6𝑥
2
5 30 +^ 𝑐
=
Let u = 𝑥^2 + 3𝑥 + 4 𝑑𝑢𝑑𝑥 = 2𝑥 + 3
𝑑𝑢 = (2𝑥 + 3)𝑑𝑥
= 𝑑𝑢𝑢
= 𝑙𝑛𝑢 + 𝑐
= 𝐥𝐧 𝒙𝟐^ + 𝟑𝒙 + 𝟒 + 𝒄
(^2) 𝑑𝑥 (𝑥 3 −1)^4
Let u = 𝑥^3 − 1 𝑑𝑢𝑑𝑥 = 3𝑥^2
𝑑𝑢 3
𝑑 𝑢 3 𝑥^4
= 13 𝑢−^4
= 13 𝑢
− 3 − 3 +^ 𝑐
= 𝑢
− 3 − 9 +^ 𝑐
= − (^) 𝟗(𝒙𝟑𝟏−𝟏)𝟑 + 𝒄
(^2) 𝑥𝑑𝑥 𝑎+𝑏 𝑡𝑎𝑛𝑥 `
Let u = 𝑎 + 𝑏 𝑡𝑎𝑛𝑥
𝑑𝑢 𝑑𝑥 =^ 𝑏^
sin 𝑥 𝑐𝑜𝑠𝑥 ;^
𝑑𝑢 𝑏 =^ 𝑠𝑒𝑐
𝑑𝑢 𝑏 𝑢
= (^1) 𝑏^ 𝑑𝑢𝑢
= 𝟏𝒃 𝐥𝐧 𝒂 + 𝒃𝒕𝒂𝒏𝒙 + 𝒄
Let u = 𝑡𝑎𝑛 3 𝑥
𝑑𝑢 𝑑𝑥 = 3𝑠𝑒𝑐
1 (^2) (𝑑𝑢 3 )
(^32) 3 ] +^ 𝑐
= 13 2tan 3𝑥
(^32) 3 +^ 𝑐
𝟑 𝟐 𝟗 +^ 𝒄
(^2) +14𝑥+ 𝑥+4 𝑑𝑥
= (^) 𝑔𝑓^ ((𝑥𝑥)) = 𝑄 𝑥 𝑑 𝑥 + 𝑅𝑔 𝑥𝑥 𝑑(𝑥)
-4 3 14 13
-12 -
3 2 5 - R(x)
𝑄 𝑥 = 3𝑥 + 2 𝑥 + 4 = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑔(𝑥)
= (3𝑥 + 2) 𝑑𝑥 + (^) 𝑥+4^5 𝑑𝑥
For the second integral :
𝑙𝑒𝑡 𝑢 = 𝑥 + 4 ;
2 2 + 2𝑥^ + 5𝑙𝑛𝑢^ +^ 𝑐]
= 𝟑𝒙
𝟐 𝟐 +^ 𝟐𝒙^ +^ 𝟓𝐥𝐧^ (𝒙^ +^ 𝟒) +^ 𝒄
(^5) − 2 𝑥 (^3) − 2 𝑥 𝑥 2 +1 𝑑𝑥 𝑥^3 − 3 𝑥 𝑥^2 + 1 𝑥^5 − 2 𝑥^3 − 2 𝑥 𝑥^5 + 𝑥^3 − 3 𝑥^3 − 2 𝑥 − 3 𝑥^3 − 3 𝑥 𝑥
𝑓(𝑥) 𝑔(𝑥)dx =^ 𝑄 𝑥 𝑑𝑥^ +^
𝑅(𝑥) 𝑔(𝑥) 𝑑𝑥
= 𝑥^3 − 3 𝑥 𝑑𝑥 + (^) 𝑥 2 𝑥+1 𝑑𝑥
4 4 −^
3 𝑥 2 2 +^
𝑥 𝑥 2 +1 𝑑𝑥
For the 2nd^ term
Let u = x^2 +
𝑑𝑢 𝑑𝑥
4 4 −^
3 𝑥 2 2 +
𝑑𝑢 2 𝑢
= 𝒙
𝟒 𝟒 −^
𝟑𝒙𝟐 𝟐 +^
𝟏 𝟐 𝐥𝐧 𝒙
11. 𝑐𝑜𝑠 cos^6 2 𝑥𝑑𝑥 (^3) 𝑥
Let u = 3x ; 2u = 6x 𝑑𝑢 𝑑𝑥 = 3 ;^
𝑑𝑢 3 =^ 𝑑𝑥
= 𝑐𝑜𝑠 2 𝑢 𝑑𝑢 3 cos 2 𝑢
= (^23) 𝑐𝑜𝑠𝑢^1 𝑐𝑜𝑠𝑢𝑐𝑜𝑠𝑢 𝑑𝑢
= 23 𝑠𝑒𝑐𝑢𝑑𝑢
= 𝟐𝟑 𝐥𝐧 𝒔𝒆𝒄𝟑𝒙 + 𝒕𝒂𝒏𝟑𝒙 + 𝒄
15. 4 sin^
(^2) 𝑥𝑐𝑜 𝑠 (^2) 𝑥 𝑠𝑖𝑛 2 𝑥𝑐𝑜𝑠 2 𝑥 𝑑𝑥
= (4𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥^ )( 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑐𝑜𝑠 2 𝑥 )𝑑𝑥
Let u = 2x 𝑑𝑢 𝑑𝑥 = 2^
𝑑𝑢 2 =^ 𝑑𝑥 𝑠𝑖𝑛𝑢 𝑐𝑜𝑠𝑢.^
𝑑𝑢 2
= 12 𝑡𝑎𝑛𝑢𝑑𝑢
Let u = 3x 𝑑𝑢 𝑑𝑥 = 3^
𝑑𝑢 3 =^ 𝑑𝑥
=
𝑑𝑢 3 𝑠𝑖𝑛𝑢𝑡𝑎𝑛𝑢
=^13 𝑐𝑠𝑐𝑢 + 𝑐
= − 𝟏𝟑 𝒄𝒔𝒄𝟑𝒙 + 𝒄
= 𝑒−^2 𝑥^ dx
𝑙𝑒𝑡 𝑢 = 2𝑥 ;
= cos 𝑢 ; 𝑑𝑣 = 𝑐𝑜𝑠𝑢𝑑𝑢
𝒔𝒊𝒏𝟒𝒙 𝟒 +^ 𝒄
3 𝑥 (^2) 𝑑𝑥
3 𝑥 (^2) + 𝑐
𝟑𝒙 𝟑 +^ 𝒄
3 − 2 𝑥 𝑙𝑛 5 +^ 𝑐
= − 𝟓
𝟑−𝟐𝒙 𝒍𝒏𝟐𝟓 +^ 𝒄
𝒙 𝒍𝒏𝟔 +^ 𝒄
1. Given slope 3 𝑥^2 + 4
𝑑𝑦 𝑑𝑥 = 3𝑥
3 3 + 4𝑥^ +^ 𝑐
𝒚 = 𝟑𝒙𝟐^ + 𝟒𝒙 + 𝒄
3. Given slope 𝑥 𝑦−+1 1
𝑑𝑦 𝑑𝑥
2 2 +^ 𝑥^ +^ 𝑐^2
𝑦^2 − 2 𝑦 = 𝑥^2 + 2𝑥 + 2𝑐
𝒙𝟐^ − 𝒚𝟐^ + 𝟐𝒚 + 𝟐𝒙 + 𝟐𝒄 = 𝟎
5. Given slope (^) 𝑥𝑦^1
𝑑𝑦 𝑑𝑥 =
𝑦 2 2 =^
ln 𝑥 2 2 +^ 𝑐^2
𝒚𝟐^ = 𝒍𝒏𝒙𝟐^ + 𝟐𝒄
7. Given slope 𝑦^
2 𝑥 ,^ through^ 1, 𝑑𝑦 𝑑𝑥 =^
− ln 𝑥 −
− ln 1 −
− ln 𝑥 −
− 4 𝑦 ln 𝑥 − 4 + 𝑦 = 0 𝟒𝒚 𝐥𝐧 𝒙 − 𝒚 + 𝟒 = 𝟎
9. Given slope 𝑦, through 1, 𝑑𝑦 𝑑𝑥 =^ 𝑦
𝑦−
1 (^2) 𝑑𝑦 = 𝑑𝑥
1 2 1 2
When 𝑥 = 1 , 𝑦 = 1 2 1 = 1 + 𝑐 ; 𝑐 = 1
2 𝑦^1 2 = 𝑥 + 𝑐
2
11. Given slope 𝑥−^2 , through 1,
𝑑𝑦 𝑑𝑥 =
𝑦 = − (^1) 𝑥 + 3 x
𝑥𝑦 = −1 + 3𝑥
𝒙𝒚 − 𝟑𝒙 + 𝟏 = 𝟎
a=-32 ft/sec^2
a=-
𝑑𝑦 𝑑𝑡 =^ −^32
𝑑𝑣 = − 32 𝑑𝑡
v=-32t+c
𝑑𝑠 𝑑𝑡
s=16t^2 + c 1 t + c 2
when t = 0, v = vo
v=-32t + c 1
vo= -32(0) + c 1
vo =c 1
v = -32t + vo when t = 1 sec, s=h=48ft h=-16t^2 + vot + c 1 48 = -16(1)^2 + vo(1) + c 2 64 - vo = c 2 When t = 0, s = 0, c 2 = 0 s = -16t^2 + vot when t = 1 sec, s = 48 s = -16t^2 + c 1 t 48 = -16(1)^2 + c 1 (1) c 1 = s=-16t^2 + 64t v = -32t + 64 @ max, v = 0 0 = -32t + 64 32t= t = 2 sec s = -16t^2 + 64t s = -16(2)^2 + 64(2) s = 64ft
1. ʃ sin 5𝑥 sin 𝑥 𝑑𝑥
= 2 sin 𝑢 sin 𝑣 𝑑𝑥
= [cos 𝑢 − 𝑥 − cos(𝑢 + 𝑣)]𝑑𝑥
𝑢 = 5𝑥 𝑣 = 𝑥
=
ʃ [cos 5 𝑥 − 𝑥 − cos (5𝑥 + 𝑥)]𝑑𝑥
2 ʃ [cos 4𝑥 −^ cos 6𝑥]𝑑𝑥
=
[ ʃ cos 4𝑥𝑑𝑥 − ʃ cos 6𝑥𝑑𝑥
= 12 [^14 sin 4𝑥 − 16 sin 6𝑥 ] + 𝐶
3. ʃ sin 9 𝑥 − 3 cos 𝑥 + 5 𝑑𝑥
= 1 2 ʃ^ [sin^ 9x^ −^ 3 + x + 5^ + sin^9 𝑥 −^3 − 𝑥 −^5 𝑑𝑥
=
2 ʃ [sin^5 𝑥^ + 2^ + sin(3𝑥 −^ 8)]𝑑𝑥 𝑙𝑒𝑡 𝑧 = 5𝑥 + 2 ; 𝑙𝑒𝑡 𝑤 = 3𝑥 − 8 𝑑𝑧 𝑑𝑥 = 5^ ;^
= 12 [− cos 𝑧 15 − 13 𝑐𝑜𝑠𝑤] + 𝐶
5. ʃ cos 3 𝑥 − 2 𝜋 cos 𝑥 + 𝜋 𝑑𝑥
=
ʃ [cos 𝑢 + 𝑣 + cos (𝑢 − 𝑣)]𝑑𝑥
𝑙𝑒𝑡 𝑢 = 3𝑥 − 2 𝜋 𝑣 = 𝑥 + 𝜋 𝑢 + 𝑣 = 3 𝑥 − 2 𝜋 + 𝑥 + 𝜋 = 4𝑥 − 𝜋 𝑢 − 𝑣 = 3 𝑥 − 2 𝜋 − 𝑥 + 𝜋 = 2𝑥 − 3 𝜋
=
2 ʃ [cos^4 𝑥 − 𝜋^ + cos(2𝑥 −^3 𝜋)]𝑑𝑥 𝑓𝑜𝑟 cos 4𝑥 − 𝜋 = cos 4𝑥𝑐𝑜𝑠𝜋 + 𝑠𝑖𝑛 4 𝑥𝑠𝑖𝑛𝜋 = −𝑐𝑜𝑠 4 𝑥 𝑓𝑜𝑟 cos 2𝑥 − 3 𝜋 = cos 2𝑥𝑐𝑜𝑠 3 𝜋 + sin 2𝑥𝑠𝑖𝑛 3 𝜋 = − cos 2𝑥
= 12 ʃ (cos 4𝑥 − cos 2𝑥)𝑑𝑥
= 12 [− 14 sin 4𝑥 − 12 sin 2𝑥] + 𝐶
= 2 ʃ [sin 8 𝑥 + 3𝑥 + 𝑠𝑖𝑛𝑥 8 𝑥 − 3 𝑥 𝑑𝑥
= 2 ʃ [𝑠𝑖𝑛 11 𝑥 + sin 5𝑥]𝑑𝑥
𝑙𝑒𝑡 𝑢 = 11𝑥 ; 𝑙𝑒𝑡 𝑣 = 5𝑥 𝑑𝑢 𝑑𝑥 = 11^ ;^
= 2[− 111 cos 11𝑥 − 15 cos 5𝑥 ] + 𝐶
= −
ʃ [cos 𝑢 − 𝑣 − cos (𝑢 + 𝑣)]𝑑𝑥
2 ʃ [𝑐𝑜𝑠^2 𝑥^ +^
𝑓𝑜𝑟 cos 2 𝑥 + 𝜋 2 = cos 2𝑥𝑐𝑜𝑠
− sin 2𝑥 sin
𝑓𝑜𝑟 cos 6 𝑥 + 𝜋 6 = 𝑐𝑜𝑠 6 𝑥𝑐𝑜𝑠
=^52 ʃ [− 𝑠𝑖𝑛 2 𝑥 − 23 𝑐𝑜𝑠 6 𝑥 +^12 𝑠𝑖𝑛 6 𝑥 ]𝑑𝑥
= 52 [^12 𝑐𝑜𝑠 2 𝑥 − 123 𝑠𝑖𝑛 6 𝑥 − 121 𝑠𝑖𝑥 6 𝑥 + 𝐶
= 𝟓𝟒 𝒄𝒐𝒔 𝟐𝒙 − 𝟓𝟐𝟒^ 𝟑 𝒔𝒊𝒏 𝟔𝒙 − (^) 𝟏𝟐𝟓 𝒔𝒊𝒏 𝟔𝒙 + 𝑪
7. ( 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)^2 dx
= (𝑠𝑖𝑛𝑥 + 2 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 + 𝑐𝑜𝑠^2 𝑥)𝑑𝑥
1 (^2) 𝑐𝑜𝑠𝑥𝑑𝑥 + 𝑐𝑜𝑠^2 𝑥𝑑𝑥
1 (^2) 𝑐𝑜𝑠𝑥𝑑𝑥 + (1+𝑐𝑜𝑠 2 2 𝑥)𝑑𝑥
Let u = sinx 𝑑𝑢 𝑑𝑥 =^ 𝑐𝑜𝑠𝑥 𝑑𝑢 = 𝑐𝑜𝑠𝑥𝑑𝑥
= 𝑠𝑖𝑛𝑥𝑑𝑥 + 2 𝑢
1 (^2) 𝑑𝑢 + 1 2 𝑑𝑥^ +^
𝑐𝑜𝑠 2 𝑥 2 𝑑𝑥
= - 𝑐𝑜𝑠𝑥 + 2 23 𝑢
3 (^2) + 𝑥 2 + 𝑠𝑖𝑛 4 2 𝑥+ 𝐶
𝟑 𝟐 (^) + 𝒙𝟐 + 𝒔𝒊𝒏𝟐𝒙𝟒 + 𝑪
3 3 +^ 𝑐
=
= 𝑠𝑖𝑛^7 𝑥 − 𝑠𝑖𝑛^9 𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 u=sinx du=cosxdx
= 𝑢^7 − 𝑢^9 𝑑𝑢
3 3 +^
𝑢^5 5 +^ 𝑐
𝟑𝟐𝒙 𝟔 +^
𝒕𝒂𝒏𝟓𝟐𝒙 𝟏𝟎 +^ 𝒄
1 2 𝑥 (^) 𝑠𝑒𝑐^4 𝑥𝑠𝑒𝑐^2 𝑥𝑑𝑥
1 (^2) 𝑥(1 + 𝑡𝑎𝑛^2 𝑥)^2 𝑠𝑒𝑐^2 𝑥𝑑𝑥
1 (^2) 𝑥(1 + 2𝑡𝑎𝑛^2 𝑥 + 𝑡𝑎𝑛^4 𝑥)𝑠𝑒𝑐^2 𝑥𝑑𝑥
1 (^2) 𝑥 + 2𝑡𝑎𝑛
5 (^2) 𝑥 + 𝑡𝑎𝑛
9 (^2) 𝑥)𝑠𝑒𝑐^2 𝑥𝑑𝑥
1 (^2) 𝑥 + 2𝑢 5 (^2) 𝑥 + 𝑢 9 (^2) 𝑥)𝑑𝑢
(^32) 3 +^
4 𝑢 (^72) 7 +^
2 𝑢 (^112) 11 +^ 𝑐
𝟑𝟐 𝒙 𝟑 +^
𝟒𝒕𝒂𝒏 𝟕𝟐 𝒙 𝟕 +^
𝟐 𝒕𝒂𝒏 𝟏𝟏𝟐 𝒙 𝟏𝟏 +^ 𝒄
5 5 +^
𝑢^7
𝟓𝒙 𝟓 +^
𝒄𝒐𝒕𝟕𝒙
4 4 𝑑𝑥 −^
𝑢^2 2 +^
1 4 𝑙𝑛 𝑐𝑜𝑠^4 𝑥^ +^ 𝑐
𝟒𝟒𝒙 𝟏𝟔 +^
𝒄𝒐𝒕𝟐𝟒𝒙 𝟖 +^
𝟏 𝟒 𝒍𝒏 𝒄𝒐𝒔𝟒𝒙^ +^ 𝒄
1 (^2 3) 𝑥 𝑐𝑠𝑐^2 3 𝑥 𝑐𝑠𝑐^2 3 𝑥 𝑑𝑥
= 𝑐𝑜𝑡
1 (^2 3) 𝑥 1 + 𝑐𝑜𝑡^2 3 𝑥 𝑐𝑠𝑐^2 3 𝑥 𝑑𝑥
= 𝑐𝑜𝑡
1 (^2 3) 𝑥 + 𝑐𝑜𝑡
5 (^2 3) 𝑥 𝑐𝑠𝑐^2 3 𝑥 𝑑𝑥
𝑙𝑒𝑡 𝑢 = 𝑐𝑜𝑡 3 𝑥 𝑑𝑢 𝑑𝑥
1 (^2) + 𝑢
5 (^2) 𝑑𝑢
(^32) 3 2
(^72) 7 2
𝟑 𝟐 (^) 𝟑𝒙 − 𝟐 𝟐𝟏 𝒄𝒐𝒕
𝟕 𝟐 (^) 𝟑𝒙 + 𝒄
(^5 2) 𝑥𝑑𝑥 𝑠𝑖𝑛 8 2 𝑥 =^
𝑐𝑜𝑠 5 2 𝑥𝑑𝑥 𝑠𝑖𝑛 5 2 𝑥
1 𝑠𝑖𝑛 3 2 𝑥 𝑑𝑥
= 𝑐𝑠𝑐^4 2 𝑥 − 2 𝑐𝑠𝑐^2 2 𝑥 + 1 𝑐𝑠𝑐^2 2 𝑥 𝑐𝑠𝑐 2 𝑥 𝑐𝑜𝑡 2 𝑥 𝑑𝑥
= 𝑐𝑠𝑐^6 2 𝑥 − 2 𝑐𝑠𝑐^4 2 𝑥 + 𝑐𝑠𝑐^2 2 𝑥 𝑐𝑠𝑐 2 𝑥 𝑐𝑜𝑡 2 𝑥 𝑑𝑥
𝑙𝑒𝑡 𝑢 = 𝑐𝑠𝑐 2 𝑥
𝑑𝑢 𝑑𝑥 =^ −^2 𝑐𝑠𝑐^2 𝑥 𝑐𝑜𝑡^2 𝑥
7 7 −^
2 𝑢^5 5 +^
𝑢^3 3 +^ 𝑐
= −
(^4) 𝑥 𝑐𝑜𝑡 6 𝑥 𝑑𝑥
= 𝑐𝑜𝑡−^6 𝑥 𝑐𝑠𝑐^2 𝑥 𝑐𝑠𝑐^2 𝑥 𝑑𝑥
− (^5) 𝑥 5 +^
𝑐𝑜𝑡 −^3 𝑥 3 +^ 𝑐
=