Integral Calculus Reviewer.pdf

3

X – 4x

nd

INTEGRAL CALCULUS REVIEWER (2 Sem 2011–2012)

∫

2. Integration – process of a function whose derivative or differential is given

-3

+ 3x + c

2

2

(x - 3)(x + 2) dx 3 x

∫ (x4 – x2 – 6)x-1/3 dx ∫ (x4 ▪ x-1/3 – x2 ▪ x-1/3 – 6 ▪ x-1/3 ) dx ∫ x11/3dx – ∫ x5/3dx – 6 ∫ x-1/3dx

Integrand – the given function Integral – the required function

11 5 1 n= 3 n=3 n =─ 3 14/3 8/3 3x 3x 2/3 – 14 8 – 9x + c

THEOREM: Two functions having the same derivatives differ at most by a constant

Proof: Let f(x) + g(x) be the function such that f’(x) = g’(x)

3/2

∫

3.

3

3x

2

-6 xx 3 3 2x

1/3

- 5

dx

*put all x’s outside radicals

∫

INDEFINITE INTEGRAL – if f(x) is a function whose derivative is f(x), the relation between the two is given by:

3

3x

Where:

3

1 2

3

1 2

3

1 2

∫ = integral sign f(x) = integrand F(x) = particular integrand c = constant of integration

∫ ∫ ∫

F(x) + c = indefinite integral of….

1. 2. 3.

POWER FORMULAS: n+1

x

∫ xndx = n + 1

+ c  if n ≠ –1

∫ x–1dx = lnx + c  if n = –1

3

3x

3

3x ▪x

3

3x

(

(

(

2/3

– 6x

1 2

 

3

–1

–1/3

– 6x

– 5)▪x

– 6x –1/6

1 n = –6

2/3

3 ▪ 3x 2

5/6

2/3

1 n = –3

3

∫ du = u + c ∫ (du + dv – dw) = ∫ du + ∫ dv – ∫ dw ∫ Rdu = ∫ du c

5/6

- 6x - 5 dx 3 2x

*bring out constant denominator and place variable denominator in the numerator

∫ f(x) dx = F(x) + c

PROPERTIES:

2/3

5/6

─

6 ▪ 6x 5

5/6

▪x

–1

–1

– 5x

dx

– 5▪x –1

–1

) dx

) dx

n = –1

 

─ 5 lnx + c

CONSTANT INTEGRATION 1.

If dy = (2X – 5)dx and y = 2 when x = –1, find y when x = 4. x -1 4 2 ? y dy = (2x-5)dx

∫ dy = ∫ (2x-5)dx

EXAMPLES:

2

1.

y + c = x – 5x + c

9 ∫ (3x – 6 x – x4 ) dx 2

∫ 3x2dx – ∫ 6x1/2dx – ∫ 9x–4dx 2 1/2 –4 3∫ x dx – 6∫ x dx – 9∫ x dx n=2

1 n=2

3

1/2

st

*Hindi pwedeng both sides may constant (c) so you have to choose which side to put 1 c

1 option: 2 y + c = x – 5x *substitute x & y to get c 2

n = -4 -3

x x x 3 ▪ 3 – 6 ▪ 3 – 9 ▪ -3 + c 2

2 + c = (-1) – 5(-1) c=4 *then substitute c & x to y + c = x2 – 5x 2

y + 4 = (4) – 5(4) y = -8

st

1 option: 2 y = x – 5x + c

b.

*substitute x & y to get c 2

2 = (-1) – 5(-1) + c c = -4 *then substitute c & x to y = x2 – 5x + c 2

y = (4) – 5(4) + (-4) y = -8 2.

2

dy 2 dx = 2x + c1 0 = 2 + c1` c1 = -2

Find the equation of the curve if the slope at pt (2,3) 2x + 1 is given by 2y - 3 . dy 2x + 1 slope = dx = 2y - 3  (2y – 3)dy = (2x + 1)dx

*substitute c1to the other equations to get the other 2 constants

14 c1= -2, c2 = -1 and c3 = 3

∫ (2y – 3)dy = ∫ (2x + 1)dx 2

*substitute these values to the last equation

2

y – 3y = x + x + c 2

3

x 14 2 y= 3 –x –x+ 3

*substitute pt (2,3) 2

3 – 3(3) = 2 + 2 + c c = -6 2 2 Equation: y – 3y = x + x – 6 (hyperbola) 3

3.

1 c1 3 = 3 + 2 + c2 + c3 16 = 3c1 + 6c2 + 6c3 dy slope = dx = -2 at x = 1. Substitute these values to the second equation. 2 -2 = 1 + c1(1) + c2 c1 + c2 = -3

dy If at any point (x,y) on a curve dx3 = 2 and (1,3) is the pt. of inflection at which the slope of the inflectional tangent line is -2, find the equation of the curve. 3 dy 3 dx = 2 2 d d y  2 dx dx  = 2 2 dy ddx2  = 2dx   2 dy ∫ ddx2  = ∫2dx 2 dy 2 dx = 2x + c1

4.

2

Find the area under the parabola y = 8 – x – 2x, above the x-axis. *complete the square 2

x + 2x + __ = 8 – y + __ 2 x + 2x + 1 = 9 – y 2 (x + 1) = –(y – 9) *it is a parabola the opens downward

V (-1,9) dA = (yA – yB)dx *dx = xLEFT – xRIGHT 2

dA = (8 – x – 2x)dx

∫dA = ∫ (8 – x2 – 2x)dx 3

x 2 A = 8x – 3 – x + c On the x-axis, y = 0.

d dy  dx dx  = 2x + c1 dy ddx  = (2x + c1)dx   dy ∫ ddx  = ∫(2x + c1)dx dy 2 dx = x + c1x + c2 2

dy = (x + c1x + c2)dx

∫dy = ∫ (x2 + c1x + c2)dx 3

2

x c1x y = 3 + 2 + c2x + c3 SYSTEM OF EQUATIONS: a. (1,3) is a point on a curve. So we substitute it to the last equation.

*substitute y in y = 8 – x2 – 2x

2

0 = 8 – x – 2x 2 x + 2x – 8 = 0 (x + 4) (x – 2) = 0 When x = -4, A = 0 *0 yung area pag x = -4 kasi hindi wala pang area na nabubuo sa point na yun. x3 *Substitute these values to A = 8x – 3 – x2 + c 3

(-4) 2 0 = 8(-4) – 3 – (-4) + c 80 c= 3 80 When x = 2, and c = 3 x3 *Substitute these values to A = 8x – 3 – x2 + c. We did this again tapos with x = 2 kasi may area nang

macocover sa point na yun .

∫ ds = ∫ (-32t + c1)dt 2

s = -16t + c1t + c2

3

2 2 80 A = 8(2) – 3 – 2 + 3 = 36 sq. units 5.

An art collector purchased for $1000 a painting by an artist whose works are currently increasing with respect to the time according to the formula du 2/3 dt = 5t + 10t + 50 where u dollars is the anticipated value of the painting in t years after its purchase. If this formula is used for the next 6 years, what is its anticipated value 4 years from now? du 2/3 dt = 5t + 10t + 50

When t = 0 and s = 0 *substitute these values to s = -16t2 + c1t + c2 2

0 = -16(0) + c1(0) + c2 c2 = 0 2

When s = -150 *Substitute s to s = -16t2 + 10t + 0. Negative yung s kasi opposite siya ng initial direction 2

-150 = -16t + 10t 2 16t – 10t – 150 = 0

∫ du = ∫ (5t2/3 + 10t + 50)dt 5/3

5t 2 u = 5 + 5t + 50t + c 3 5/3 2 u = 3t + 5t + 50t + c u 1000 ? 0 4 t

*get t by using the quadratic formula

t = 3.4 seconds *you will get 2 answers here. ‘yung isa negative. Siyempre, ineneglect natin ‘yung negative dahil bawal maging nega ‘yung time 

ds 2 Differentiate s = -16t + 10t to get dt /the velocity

When u = 1000 and t = 0.

ds dt = -32t + 10

*Substitute to u = 3t5/3 + 5t2 + 50t + c

c = 1000 When t = 4 and c = 1000 *Substitute to u = 3t

5/3

When t = 3.4. v = -32(3.4) + 10 v = -98.8 ft/s

2

+ 5t + 50t + c

u = $1,286.89 6.

A woman in a hot air balloon dropped her binoculars 150ft above the ground and is rising at the rate of 10ft/s. (a) How long will it take the binoculars to reach the ground? (b) With what speed will it strike the ground? dv 2 a = g = dt = -32ft/s *negative yung acceleration/gravity kasi opposite siya ng direction ng velocity ng hot air balloon

*Again, it is negative kasi opposite siya nung initial direction

DEFINITE INTEGRAL PROPERTIES: 1.

b b f(x)dx = f(x)dx a a

- interchanging the limits changes the sign of the integral 2.

b f(x)dx = a

∫ dv = ∫ -32dt v = -32t + c1

When t = 0 and v = 10ft/s *substitute these values to v = -32t + c1

s = vot + 2 at ds v = dt dv a = g = dt

3.

b f(x)dx = a

c b f(t)dt + f(z)dz a c

- The definite integral of an integrand is independent of the variable of integrations EXAMPLES: 1.

2 -1 3

ds v = dt = -32t + c1

c b f(x)dx + f(x)dx a c

- The interval of integration may be broken down to any number of subintervals and the integration performed over each interval separately

dv dt = -32

10 = -32(0) + c1 c1= 10

*Substitute c1 and c2 to s = -16t2 + c1t + c2

s = -16t + 10t + 0

5x2 + 1 x – 1  dx 3 2 

5x 1 2 1 3 +6x –2x

]

2

−1

-3

*substitute 2 and -1 sa mga x. Subtract the lower number from the upper number.

1 u 2 ▪ -3 + c 1 ─6(2x - 7)3 + c

5 3 1 2 1 3 2 3 (2 – (-1) ) + 6 (2 – (-1) ) - 2 (2+1) 5 1 1 28 3 (8 + 1) + 6 (4 – 1) - 2 (2+1) = 2 = 14 3 1 x +1 dx 0 x+1 2 1 (x + 1)(x - 2x + 1) x+1 0

2.

1 0

3.

∫

∫ (4t2 + 9)-1/2tdt 2

u = 4t + 9 du = 8tdt du 8 = tdt 1 n = -2

2

(x – 2x + 1)dx

3

1 x 2 ─ x + x 2 0 1 3 5 2 3 (1 – 0) – (1 – 0) + (1 – 0) = 6

]

* diba sa orig na formula it’s (4t2 + 9)-1/2tdt so diba u = (4t2 + 9)-1/2 tapos after that yung tdt. trinanspose natin yung 8 to the other side to follow the general formula na undu. Diba nakuha nating du nung una is 8tdt. Para maging tdt lang which is yung nasa original formula, linipat yung 8. Pero gagawin siyang constant or “preparation” sa integration.

1 -1/2 8 ∫u du 1/2 1 u ▪ 8 1 +c

THE GENERAL POWER FORMULA

∫u du n

 if n ≠ -1: u

n+1

∫undu = n + 1

tdt 2 4t + 9

2

+c

1 2 4 4t + 9 + c

 if n = -1:

∫undu = lnu + c 1.

∫ (x + 1) dx 2

4.

u=x+1 du = dx

∫ u2du

∫

1 -2 2 ∫ u du -1 1 u ▪ 2 -1 + c 1 ─ 2e2t + 1 + c

∫ (2x – 7)-4dx

*trinanspose yung 2 sa other side para maging equal yung value ng du sa original formula. Pero yung 2 na trinanspose aka yung 1 1 2 , gagawin mong constant. so if you like, hide 2 , du = dx. So it still follows the original formula na ∫und. Pag hindi ‘to nagets explain ko sa other examples. =)))

1 2

∫ u-4du

2t

e dt 2t 2 (e + 1) n = -2 u = e2t + 1 du = 2e2tdt du 2 = e2tdt

dx 4 (2x - 7) u = 2x – 7 du = 2dx du 2 = dx

2t

e dt 2t e +2 +1 4t

∫ (e2t + 1)-2e2tdt

3

u 3 +c 3 (x + 1) 3 +c 1 3 2 3 (x + 3x + 3x + 1) + c 1 3 2 1 3x +x +x+3+c

2.

∫ ∫

5.

∫

y

(y

1/3

4/3

2 + 9) dy

∫ (y4/3 + 9)-2y1/3dy u=y

4/3

+9 4 1/3 du = 3 y dy 3 1/3 4 du = y dy

3 -2 4 ∫ u du

9.

3 ─ 4(y4/3 + 9) + c

∫ (1 + 2e3x)e3xdx 3x

u = 1 + 2e 3x du = 6e dx du 3x 6 = e dx

2 2 ─ u2 = ─ 1 2 (4 x - 1)

1 1 6 ∫ u du 2 1 u ▪ 6 2 +c 3x 2 (1 + 2e ) +c 12

7.

∫

x

 

EXAMPLES when

+9 1/4 dx x 1 n=2 3/4 u=x +9 3 -1/4 du = 4 x dx 4 dx 1/4 3 du = x

1.

∫

2.

∫

4 1/2 3 ∫ u du 4 2 2/3 3▪3u +c 8 3/4 2/3 9 (x + 9) + c 8.

10

]

8

2 2 10 ─ 1 + 1 = 9 2 2 (4 (10) - 1) (4 (8) - 1)

3/4

∫ (6cos x + sin x) 2

1 x - 1 -3dx 4 

n = -3 1 u=4x–1 1 du = 4 dx 4du = dx -3 4 u du -2 4u -2

-1

3 u 4 ▪ -1 + c

6.

10 8

2

1/2

sinxcosx dx 1 n=2 2 2 u = 6cos x + sin x d(cosx) d(sinx) du = 6[2cosx ▪ dx ] + 2cosx ▪ dx du = 6 [2cosx(-sinx)]dx + 2sinxcosxdx du = -12sinxcosxdx + 2sinxcosxdx du = -10sinxcosxdx du ─10 = sinxcosxdx

1 2 2 3/2 ─15 (6cos x + sin x) + c

sec5θtan5θ 3 + 2sec5θ dθ u = 3 + 2sec5θ du = 2(5)sec5θtan5θ dθ du 10 = sec5θtan5θ dθ 1 du 10 ∫ u 1 10 lnu + c 1 10 ln(3 + 2sec5θ) + c dx x+ x *factor x +

x for it to be

∫

dx x ( x + 1) u= x+1 dx du = 2 x dx 2du = x du 2 u = 2lnu + c 2ln( x + 1) + c

∫

1 1/2 ─10 ∫ u du 1 2 3/2 ─10 ▪ 3 u + c

 du -1  : or u du = lnu + c u 

∫

3.

∫ secxdx secx + tanx ∫ secxdx ▪ secx + tanx

x ( x + 1)

2

∫

∫

EXPONENTIAL FUNCTION

secxtanx + sec x secx + tanx dx u = secx + tanx 2 du = (secxtanx + sec x)dx du u = ln(secx + tanx) + c

1

∫ audu = lna au + c ∫ eudu = eu + c TRIGONOMETRIC FUNCTIONS

4.

x

e dx x 1 + 3e

ln2 0

2.

x

u = 1 + 3e x du = 3e dx du x 3 = e dx

3.

= – ln cos u + c x

= – ln csc u + c

ln2

equation ng u which is 1 + 3e . when x = ln2, e = 2. So 1 + 3(2) = 7. And when x = 0, it’s going to be e0 = 1. So 1 + 3(1) = 4.

ln2 7

u

5.

0 4

6. 7.

1 du 3 u 1 x 3 ln(1 + 3e ) 1 3 [ln7 – ln4] 1 7 3 ln4

∫

8.

]

5.

∫

7

9.

4

10.

1.

∫ sin4xdx u = 4x du = 4dx du 4 = dx

3

x - 2x + 5 x - 3 dx

1 4 ∫ sin u du 1 4 (-cos u du) + c 1 ─ 4 cos4x + c

26

∫ (x2 + 3x + 7 + x - 3 )dx 3

2

x 3x -1 3 + 2 + 7x + 26∫ (x – 3) dx 3 2 x 3x + 3 2 + 7x + 26ln(x – 3) -2 -3

∫ sec u du = ln(sec u + tan u ) + c ∫ csc u du = ln(csc u – cot u) + c ∫ sec2u du = tan u + c ∫ csc2u du = –cot u + c ∫ sec u tan u du = sec u + c ∫ csc u cot u du = –csc u + c

EXAMPLES:

*when the degree/exponent of the numerator is higher than the denominator, divide.

6.

∫ cot u du = ln sin u + c

4.

*change the limits. To do that, substitute sa limits sa mga x sa x

∫ sin u du = – cos u + c ∫ cos u du = sin u + c ∫ tan u du = ln sec u + c

1.

2.

y+2 2 y + 4y dy

du =

2

u

x 1

2 x dx 2du = x

dx

2∫ tan u du = 2lnsec u + c

*change limits

-3 -3

tan x dx x u=

u = y + 4y du = (2y + 4)dy du 2 = (y + 2)dy

x

∫

2ln(sec x ) + c

-2 -4

1 -2 du 1 2 -3 u = 2 ln |u| 1 2 [ ln|-4| - ln|-3|] 1 4 2 ln3

3.

]

−4 −3

∫ e2xcos e2x dx u = e2x du = 2e2x dx 2du = e2x dx

1 2 ∫ cos u du 1 2x 2 sin e + c

u = tany 2 du = sec y

TRIGONOMETRIC TRANSFORMATIONS I.

∫ sinmx cosnx dx where m or n is a positive odd integer tools: change the one w/ odd powers 2 2 sin x = 1 – cos x 2 2 cos x = 1 – sin x

Ex:

∫ sin52x cos42x dx

*integrate each term. so their n’s sa un would be 4, 6, and 8 respectively. u = cosy du = -siny dy n+1 u n So we’ll be using the form ∫u = n + 1 for each term. And substitute 2x to y na ulit.

1 4 6 8 2 ∫ (u – 2u + u ) siny dy 5 7 9 1 cos 2x 2cos 2x cos 2x  –2  5 – – 7 9 +c  2 4 1 1 2cos 2x cos 2x  5 –2 cos 2x5 – – 7 9 +c  II. ∫ sec x tan x dx or ∫ csc x cot x dx a. Where m is positive even integer 2 2 tools: sec x = 1 + tan x 2 2 csc x = 1 + cot x Ex: 1 1 ∫ tan42 x sec42 x dx n

m

1 y=2x 2dy = dx

2∫ tan y sec y dy 4

2∫ tan y sec y sec y dy 4

2

2

2∫ tan y (1 + tan x) sec y dy 4

y

2

2∫ (tan y + tan y) sec y dy 4

b. Where n is a positive odd integer 2 2 tools: tan x = sec x – 1 2 2 cot x = csc x – 1 Ex: y = 3x dy 3 = dx

1 5 4 2 ∫ sin y cos y dy 1 4 4 2 ∫ sin y cos y siny dy 1 2 2 4 2 ∫ (sin ) cos y siny dy 1 2 2 4 2 ∫ (1 – cos y) cos y siny dy 1 2 4 4 2 ∫ (1 – 2cos y + cos y) cos y siny dy 1 4 6 8 2 ∫ (cos y – 2cos y + cos y) siny dy

4

6

∫ tan53x sec33x dx

y = 2x dy = 2dx dy 2 = dx

m

tan512 x tan712 x  2∫ (u + u )du = 2 5 + 7  +c   4

6

2

n

1 5 3 3 ∫ tan y sec y dy 1 4 2 3 ∫ tan y sec y tany secy dy 1 2 2 2 3 ∫ (sec y – 1) sec y tany secy dy 1 4 2 2 3 ∫ (sec y – 2sec y + 1) sec y tany secy dy 1 6 4 2 3 ∫ (sec y – 2sec y + sec y) tany secy dy u = secy du = tany secy dy

1 6 4 2 3 ∫ (u – 2u + u )du 7 5 3 1 sec 3x 2sec 3x sec 3x  – + 3 7 5 3 +c

∫  ∫ ∫

1 + sinθ  2 cosθ  dθ 2

1 + 2sinθ + sin θ 2 dθ cos θ 1 2 cos θ dθ +

∫

∫

2

2sinθ sin θ 2 2 cos θ dθ + cos θ dθ sinθ ∫ sec2θ dθ + 2∫ secθ ▪ cosθ dθ + ∫ tan2θ dθ 2tan θ + 2∫ secθ tanθ dθ + ∫ (sec θ – 1) dθ 2

2

2tan θ + 2secθ + ∫ sec θ dθ – ∫dθ 2 2tan θ + 2secθ + tanθ – θ + c 2

2

III. ∫ tan x dx or cot x dx where n is an integer 2 2 tools: tan x = sec x – 1 2 2 cot x = csc x – 1 a. n is a positive even integer EX: n

n

∫ tan6x dx ∫ tan4x ▪ tan2x dx ∫ tan4x (sec2x – 1) dx

∫ (tan4x sec2x – tan4x) dx

IV. ∫ sin x cos x dx where m & n are positive even integers m

*step by step nating i-solve each part, okay? So we’ll start 4 2 with ∫tan x sec x

tools:

∫ tan x sec x dx 4

n

2

u = tanx 2 du = sec x dx 5

y = 3x dy = 3dx dy 3 = dx

–∫ tan x dx 4

–∫ tan x ▪ tan x dx 2

2

–∫ tan x(sec x – 1) dx 2

–∫ (tan x sec x – tan x) dx 2

2

2

2

n

*it’s possible na to integrate tan x sec x. Use ∫u du. And distribute the negative sign so magiging positive yung 2 tan x. 3

5

+ ∫ tan x dx 2

+ ∫ (sec x – 1) dx 2

+ ∫ sec x dx – ∫dx 2

+ tan x – x

*combine na the two parts. So the final answer would be: 3

tan x tan x 5 – 3 + tanx – x + c

∫ tan5x dx ∫ tan3x ▪ tan2x dx ∫ tan3x(sec2x – 1) dx ∫ tan3x sec2x dx – ∫ tan3x dx

n

*pwede na ma-integrate yung first term using u du so 3 we’ll focus on the second term which is tan xdx

tan x 2 4 – ∫ tan x ▪ tanx dx 4 tan x 2 4 – ∫ tanx(sec x – 1) dx 4 tan x 2 4 – ∫ (tanx sec x – tanx) dx 4 tan x 2 4 – ∫ (tanx sec x) dx – ∫ tanx dx 4 2 tan x tan x – 4 2 – ln(secx) + c

1 2 2 3 ∫ sin y cos y dx 1 2 3 ∫ (siny cosy) dx 1 1 2 3 ∫ (4 sin 2y)dy 1 2 12 ∫ sin 2ydy 1 1 12 ∫ 2 (1 – cos4y)dy 1 1 24 ∫ dy – 24 ∫ cos4ydy 1 1 24 [y – 4 sin4y] + c 1 1 24 [3x – 4 sin12x] + c

∫ sin2x cos4x dx ∫ sin2x cos2x cos2x dx ∫ (sinx cosx)2 cos2x dx 1 ∫ (2 sin2x)2cos2x dx

b. n is a positive odd integer EX:

4

1

∫ sin23x cos23x dx

*next is –∫ tan x dx

tan x – 3 3 tan x – 3 3 tan x – 3 3 tan x – 3

2

Ex: 4

2

1

cos x = 2 (1 + cos2x)

u tan x 5 = 5

2

2

sin x = 2 (1 – cos2x)

∫ u4du 5

1

sinx cosx = 2 sin2x

1 2 2 4 ∫ sin 2x cos x dx 1 2 8 ∫ (1 – cos4x) cos x dx 1 1 2 2 8 ∫ cos x dx – 8 ∫ cos4x cos x dx 1 1 1 1 8 ∫ 2 (1 + cos2x) dx – 8 ∫ cos4x ▪ 2 (1 + cos2x) dx 1 1 ∫ (1 + cos2x) dx – 16 16 ∫ (cos4x + cos2x cos4x) dx 1 1 1 1 16 (x + 2 sin2x) – 16 ∫ cos4x dx – 16 ∫ (cos2x cos4x) dx 1 1 1 1 1 2 16 (x + 2 sin2x) – 16 ▪ 4 sin4x – 16 ∫ (cos2x(1 – 2sin 2x) dx 1 1 1 1 2 16 (x + 2 sin2x) – 64 sin4x – 16 ∫ (cos2x – 2sin 2x cos2x) dx 1 1 1 1 1 2 16 (x + 2 sin2x) – 64 sin4x – 16 ∫ cos2x dx – 16 ▪ 2∫ sin 2x *use u = sin2x, du = 2cos2x dx cos2x dx

1 1 1 1 1 1 1 2 16 (x + 2 sin2x) – 64 sin4x – 16 ▪ 2 sin2x – 8 ▪ 2 ∫ u du 3 1 1 1 1 1 u x + sin2x – sin4x – sin2x – ▪ 16 32 64 32 16 3 + c 3 1 1 1 1 1 u x + sin2x – sin4x – sin2x – ▪ 16 32 64 32 16 3 + c 1 1 1 1 1 3 16 x + 32 sin2x – 64 sin4x – 32 sin2x – 48 sin 2x + c

V.∫ sin ax sin bx dx

∫ sinmx cosnx dx

1 2 1 2

[ 12 sin2(π4 ) + 41 sin4(π4 ) – 21 sin2(0) + 14 sin4(0) ] [ 12 (1) + 14 (0)] – 0 = 41 π 3

4.

0

sinx sin2x sin3x dx

1 2 ∫ sinx[cos(2x – 3x) – cos(2x + 3x)] dx 1 2 ∫ sinx[cosx – cos5x] dx 1 2 ∫ (sinx cosx – sinx cos5x) dx *for sinx cosx, u = sinx, du = cosxdx. So the formula you’ll use would be ∫undu.

∫ sinmx cosnx dx 2

tools:

1 sinα sinβ = 2 *cos(α ─ β) – cos(α + β)] 1 cosα cosβ = 2 *cos(α ─ β) + cos(α + β)+ 1 sinα cosβ = 2 *sin(α ─ β) + sin(α + β)+

EX: 1.

∫ sin4x sin7x dx 1 2 ∫ [cos(4x – 7x) – cos(4x + 7x)] dx 1 2 ∫ [cos(–3x) – cos(11x)] dx 1 2 ∫ (cos3x – cos11x) dx 1 1 1 2 [ 3 sin3x – 11 sin11x] + c

2.

∫ cos7x sin4x dx 1 2 ∫ [sin(4x – 7x) + sin(4x + 7x)] dx 1 2 ∫ [sin(–3x) + sin(11x)] dx 1 2 ∫ (–sin3x + sin11x) dx 1 1 1 2 [3 cos3x – 11 cos11x] + c π 4

0

cosx cos3x dx

1 2 ∫ [cos(x – 3x) + cos(x + 3x)] dx 1 2 ∫ (cos2x + cos4x) dx 1 1 1 2 2 sin2x + 4 sin4x

[

]

2

sin x 1 1 4 + 16 cos4x – 24 cos6x

]

π 3

0

9 = 32

INVERSE TRIGONOMETRIC FUNCTIONS 1.

2. 3.

∫ ∫ ∫

du -1 u 2 2 = Sin a + c a ─u du 1 -1u 2 2 a + u = a Tan a + c du 1 -1 u 2 2 = a Sec a + c u u ─a

Examples:

*let α = 4x and β = 7x

3.

1 sin x 1 1 2 ▪ 2 – 2 ∫ 2 [sin(x – 5x) + sin(x + 5x)] dx 2 sin x 1 4 – 4 ∫ (-sin4x + sin6x) dx 2 sin x 1 1 4 + 4 ∫ sin4x dx – 4 ∫ sin6x dx

π 4

0

1.

∫ ∫ ∫

dx 2 25 + 64x dx 2 2 (5) + (8x)

1 du 2 2 8 a +u 1 1 -1 8x 8 ▪ 5 Tan 5 + c 1 -1 8x 40 Tan 5 + c

a=5

u = 8x

du 8 = dx

2.

∫ ∫ ∫

∫

dx 2 9 ─ 4x dx 2 2 (3) ─ (2x)

a=3

u =2x

du 2 2 a ─u 1 -1 3x + 1 Sin  4  + c   3

du 2 = dx

1 2

du 2 2 a ─u 1 -12x 2 Sin 3 + c

3.

∫ ∫ ∫ ∫

6.

2

sec x dx 2 50 ─ sec x

∫

2

sec x dx 2 50 ─ (1 + tan x)

∫

2

sec x dx 2 49 ─ tan x

∫

a = 7 u =tanx

du = sec2x dx

2

*add and subtract 4 para maging perfect square yung x – 4x

7.

∫

dx 2 25 ─ (x ─ 2) -1 x ─ 2 Sin  5  + c  

∫

dx 4 4 (5x - 4x + 5 ) + (2 - 5 ) 2

dx ( 5x─ du 2 2 a +u

a=5

∫ ∫ ∫

u=x–2

du = dx

1 1 2 5 + 3 ─ (3x + 2x + 3 ) dx

 4

3

 2 ─ ( 3 x + 1 )2 3  u =

3x+

5

  

2 5

u=

5x─

2 5

= dx

5x - 2 5 6 5

 +c 

dx 2 x 9x - 25

3dx 2 2 3x (3x) - (5)

1

dx

4 3

du

a=

0

(x + 1)dx 2 x +1

∫

x 2 x + 1 dx +

∫

dx 1 2 x +1

x *For x2 + 1 : du 2 = xdx

u = x2 + 1

∫ ∫ du u +

dx 1 2 x +1

dx *For x1 + 12 : u=x

1 3

a=5

du 2 2 u u ─a 1 -13x 5 Sec 5 + c 8.

22 1 formula c = 4(3) = 3

∫

2 5  6

yung x na nasa baba.

dx 2 5 - 2x - 3x

∫

 

2 2 ) + 5

3 *multiply the whole equation to 3 para maging 3x

b2 *to get c in ax2 + bx + c, get the value of 4a . so in this

a=

dx 5x - 4x + 2

dx 2 21 + 4 ─ (x ─ 4x + 4) *nakalagay sa equation, + 4 sa pareho, kasi yung second na 4, negative siya pag dinitribute yung nega

5.

= dx

1 -15x - 2 Tan +c 6 6

dx 2 21 - 4x + x

∫

3

2

1 5 -1 ▪ Tan 5 6

2

sec x dx 2 2 (7) ─ (tanx) -1 tanx Sin  7  + c  

4.

∫ ∫

du

1 2

∫ ∫ du u +

a=1

du 2 2 u +a

du = dx

u = 3x

du 3 = dx

1 1 2 -1 ln(x + 1) + Tan x 2 0 1 2 2 -1 -1 2 [(ln1 + 1) – (ln0 +1)] + [Tan 1 – Tan 0] 1 π 2 [ln2 – ln1] + [4 - 0] 1 π 2 ln2 + 4

du = 2x – 4

]

e

9.

∫

∫

5 dx 2 2 2 2 ln(x – 4x + 20) + 8 (x - 2) + (4) a=4 u = x – 2 du = dx

∫

5 du 2 2 2 2 ln(x – 4x + 20) + 8 a + u 5 1 2 -1x - 2  2 ln(x – 4x + 20) + 8 ▪ 4 Tan  4  5 -1 -1 2 [ln32 – ln16] + 2 [Tan 1 – Tan 0] 5 32 π 2 ln16 + 2 ▪ 4 5 π 2 ln2 + 2

dy 2 y(1 + ln y)

1

∫ ∫

dy 2 2 y(1 + (lny) )

a=1

dy du = y

u = lny

du 2 2 a +u

]

-1

e

Tan (lny) 1 -1 -1 Tan (lne) – Tan (ln1) -1 -1 Tan 1 – Tan 0 π 4 4

10.

2

12.

∫

3 ─2 (-2x + 2) + 4

∫

3 ─2

dy 2

y ( y) ─1

2

y . So yung y sa labas, ihiwalay mo para

3 ─2

1 maging

. Para yung form maging

y

u=

y

2du =

du u u2 ─ a2

-1

-1

2 [Sec

y

]

y

∫

2 -1

4 ─ Sec

2]

13. 2

*let u = x – 4x + 2z and du = (2x – 4)dx *divide the numerator by the derivative of the denominator. Then follow this: 5 2 (2x - 4) + 8

∫ ∫

∫

(2x - 4)dx dx 2 x - 4x + 200 + 8 x - 4x + 200 2

2

u = x – 4x + 200

1

du = (2 – 2x)dx

u

∫

dx 2 3 + 2x - x

du = (2 – 2x)dx

-1/2

du + 4

∫

∫

dx 2 (3 + 1) - (x - 2x + 1)

dx 2 (2) - (x - 1) 2

∫

]

2 1

dx (x + 1) 2x(x + 2)

∫

dx *linabas lang yung 2 (x + 1)( 2 ) x(x + 2)

1 2

2

x - 4x + 20

5 2

2

dx

du 2 2 a ─u 2 -1 x - 1 – 3 3 + 2x - x + 4 Sin  2    π *─3 3 – (-6)] + 4[ 6 ─ 0+ 2π 6-3 3+ 3

4

(5x - 2)dx 2 x - 4x + 20

2

2

2

2

11.

6

– 3 3 + 2x - x + 4

π π π 2 [ 3 ─ 4 ] = 12

6

u = 3 + 2x – x2

-2x + 2 2 dx + 4 3 + 2x - x

1/2

du 2 2 u u ─a

2 Sec

∫

3 2u ─2 ▪ 1 + 4

dy

∫

2

3 + 2x - x

u = 3 + 2x – x2

*Diba u = y

(3x + 1)dx 2 3 + 2x - x

1

]

*divide the numerator by du. Yung ginawa sa previous numbahhh

dy y y─1

2

∫

5 du dx 2 2 u + 8 (x - 4x + 4) + (20 - 4)

1 2

∫ ∫

dx 2 (x + 1) x + 2x + 1 - 1 dx 2 2 u=x+1 (x + 1) (x + 1) - 1

du = dx

a=1

∫

∫

1 du 2 2 2 u u ─a 2 1 -1 Sec (x + 1) 1 2 1 -1 -1 [Sec 3 – Sec 2] 2 1 π -1 [Sec 3 – 3 ] 2

(x

]

∫

3.

∫

∫ ∫

4. 5.

| uu +- aa | + c

du 1 2 2 a - u = 2a ln

| uu +- aa | + c

1 6

1 7 ln

| xx +- 52 | + c

0

∫ ∫ ∫

a=1

∫ ∫

dx 2 x - 3x - 10

2

0

2

25 - x dx

3

∫

2

a=5

u=x

du = dx

2

a ─ u du

4 1 2 -1 x  { x 25 x + 25 Sin } 2 5  3 1 4 -1  -13  2 { [12 + 25 Sin 5  ] – [12 + 25 Sin 5  25 -14  -13  2 [ Sin 5  – Sin 5  ]

]

1

1 a=x+2

du 2 2 u ±a 1 2 ln|(x + 2 + x + x + 1 | + c

3.

a=

1

ln

4

5.

x + x + 4 + (1 - 4 ) dx

u=x

x+ 2 | |] 2 2 x- 2 1 1+ 2 0+ 2 [ ln| | ─ ln| |] 2 2 1- 2 0- 2 1 1+ 2 [ ln| |─ |ln1| ] 2 2 1- 2 1 1+ 2 ln| | 2 2 1- 2

dx

1 3 (x + 2 )2 + ( 2 )2

du = dx

|+c

du 2 2 a -u

dx 2 x +x+1 1

7 a=2

dx 2 2-x

dx 2 2 ( 2) -x

du 2 2 u ±a

2

3 7 2)+2

∫ ∫

1 2 4 6 ln|3x + 9x - 1 | + c

2.

7

| (x -

1

xdx 4 9x - 1 du 6 = xdx

3

(x - 2 ) - 2

1 7 ln

4.

EXAMPLES:

xdx 2 2 2 2 u = 3x (3x ) - 1

du = dx

du 2 2 u -a

du 1 2 u - a = 2a ln

∫ ∫ ∫

3 u=x-2

∫

1

1.

dx 3 2 49 2) - 4

(x -

du 2 2 u ±a |} + c 2 2 = ln|u + u ±a 1 2 2 2 2 2 -1 u a ─ u du = 2 { u a ─ u + a Sin  a  } + c   2

9 9 - 3x + 4 ) - (10 + 4 )

∫

ADDITIONAL FORMULAS: 1 2 2 2 2 2 2 2 1. ∫ u ± a du = 2 { u u ±a ± a ln |u + u ±a |} + c 2.

dx 2

du = dx

3 a= 2

*diba yung notation na “Sin-1(x)” means ANGLE yung π value niya? Like, “Sin-1(1)” = 90 or 2 . So let’s represent 4 5

-1 3

( ) = θ and Sin ( ) = β.

Sin-1

5

25 2 (θ – β) *recall the identity sin(θ – β) = sinθcosβ – cosθsinβ. Draw 4 ka ng triangle of each angle. Since sin function yung 5 at 3 opposite 5 , ile-label mo yung numbers na yan sa hypotenuse . So the triangles would look like this:

sin(θ – β) = sinθcosβ – cosθsinβ 4 4 3 3 sin(θ – β) = 5  5  – 5  5         7 sin(θ – β) = 25 *note that we’re only getting (θ – β) -1 7 (θ – β) = Sin 25    25 7 -1  2 Sin 25  π 4

6.

cos2x

π 12

1 2

1

2

sin 2x - 16

dx

1 a=4

8.

∫ ∫ ∫

du = (2x + 1)dx

(2x - 1) - 4 dx 2 x +x+2 (2x - 1) dx ─ 4 2 x +x+2

u

du u = sin2x 2 = cos2xdx

-1/2

du ─ 4

2

2(x + x + 2)

∫

∫

∫

dx 1

2

1

(x + x + 4 ) + (2 - 4 )

du 2 2 u -a 1

– 4 ln|(x + 2 ) + x + x + 2 |} + c

1/2

2

1

2

sin2x - 4

| sin2x + | ] 1 4

π 4

e^3

9.

π 12

π π 1 1 sin2 4  - 4 sin212  - 4     ln ─ ln π π 1 1 sin24  + 4 sin212  + 4     π π 1 1 sin2  - 4 sin6  - 4     ln ─ ln π π 1 1 sin 2  + 4 sin 6  + 4      34   14  ln 5  ─  3      4  4  3 1 ln5  ─ ln3     

|

| |

|

| |

∫

6y + 1 2 9y - 6y - 3 dy

|

10.

∫

du du 2 2 u +3 u -a

∫

1 3 2 3 ln(9y – 6y – 3) + 4 ln

du u = 3y + 1 3 = dy

| 3y3y +- 31 | + c

du lnx 2 = x dx

a=1

∫

du 2 2 u -a 2

e^3

2

𝑒^2

2 3

2 2

2 3

2 2

3

2

3

2

∫ ∫ ∫

1 1 + x dx x+1 2 dx ( x) 1 x

2

2

( x ) + 1 dx

u=

2

2

HYPERBOLIC FUNCTIONS 1. a=2

x

2du =

2 ∫ u ± a du 1 2 { x ▪ x + 1 + ln | x + x + 1 |} + c

2 9y - 6y - 3 dy

1 3

u = ln2x

1 4 3 1 4 4 [ ln5 ─ ln5 ] = 4 ln3

*divide the numerator by du. Yung method na ginawa before 1 3 (18y - 6) + 3

(18y - 6)dy dy 2 2 9y - 6y - 3 + 3 (9y - 6y + 1) - (3 + 1)

2

| lnln xx +- 11 |] 1 ln e - 1 ln e - 1    4 ln | ln e + 1 |─ ln | ln e + 1 | 1 3 -1 2 -1 4 [ ln| 3 + 1 |─ ln | 2 + 1 |]

du = (18y – 6)dy

1 3

lnx 4 x(ln x - 1) dx

1 1 2 ▪ 2 ln

|

u = 9y2 – 6y – 3

e^2

1 2

9 ln5

∫ ∫ ∫

u = x2 + x + 2

2 x + x + 2 ─ 4 ln|(x + 2 ) + x + x + 2 |} + c 1

7.

2x - 3 dx 2 x +x+2

*divide the numerator by du. Yung method na ginawa before

du 2 2 u -a

1 1 2 ▪ 1 ln 2(4 )

∫

2. 3. 4. 5. 6. 7. 8.

∫ sinh u du = cosh u + c ∫ cosh u du = sinh u + c ∫ tanh u du = ln |cosh u | +c ∫ coth u du = ln |sinh u | +c ∫ sech2 u du = tanh u + c ∫ csch2u du = –coth u + c ∫ sech u tanh u du = –sech u + c ∫ csch u coth u du = –csch u + c

dx x

a=1

*so diba infinity over infinity, so bawal yun. Babalik tayo sa equation before this. Yung may b over b + 16. Derive that.

EXAMPLE:

∫

(sech 1 - t )(tanh 1 - t ) dx 1-t u=

1-t

1   8 ln

─2du =

| 11 |─ ln | 171 | *recall that ln1 = 0

dt

1 1 ─ 8 ln 17

1-t

∫

a *recall that lnb = lna - lnb

─2 sech u ▪ tanh u ▪ du

1 ─ 8 [ln1 – ln17]

─2(–sech u) + c

1 1 ─ 8 [–ln17] = 8 [ln17]

2sech 1 - t + c IMPROPER INTEGRALS I. Integrals with infinite limits in the integrand

∞

2.

a b

-∞

f(x)dx

a

f(x)dx = lima-∞

a

a

1

f(x)dx

2dy y(y + 16) b

limb∞ b 1

1

u=y+8

∫

du 2 2 u -a

| yy ++ 88 +- 88 | 1 y 2 ▪ 2(8) ln | y + 16 |]  1  b 1   8 ln| b + 16 |─ ln | 1 + 16 |  1  ∞ 1    8 ln| ∞ |─ ln | 17 | 1 2 ▪ 2(8) ln

b

1

du ─ 2 = xdx

u = ─x2

u

e du b

1 1 ─2 ▪ ─1 = 2 3.

∞ 2

dx 2 x +1 b

limb∞

1

du 2 2 u +a

| | = 21 ln| xx +- 11 |]  1 b-1 2-1   2 ln| b + 1 |─ ln | 2 + 1 |  1 2a ln

u-a u+a

b 2

*so diba infinity over infinity, so derive the numerator and the denominator

2dy 2 y + 16y + 64 - 64

dy 2 2 (y + 8) - (8) a = 8

dx

0  1 1 1 ─2  eb^2 ─ e0    1 1 1 ─2 ∞ ─ 1  

b

EXAMPLES: ∞

1

]

f(x)dx

f(x)dx = lima-∞ and b∞

b

-x^2

xe

1

1 1 ─2 ex^2 

NOTE: ∞ 0 ∞ & 0 = ‘pag ganyan yung situation, dun sa equation/s kung sa’n naka substitute yung “b” or “a”, derive both the numerator and the denominator. Then you may start dividing 1 ∞=0

2

1 ─2

b

-∞ ∞

1.

limb∞

b

f(x)dx = limb∞

dx b

*in other words, isa or both a and b sa formula b na a f(x)dx, infinity. ∞

-x^2

xe

0

du = dy

1   2 ln

| 11 |─ ln | 13 | 1  1    ─ ln | 2  3 | 1 1 1 1 ─ 2 ln 3 = ─ 2 [ln1 – ln3] = 2 ln3

II. Integrals with infinite discontinuities in the integrand *in other words, isa or both a and b sa formula b

na a f(x)dx, pag sinubstitute sa f(x)dx, UNDEFINED yung lalabas.

a) If f(x) increases numerically without limit as x  a, then n m

n

f(x)dx = limam+

f(x)dx

a

a) If f(x) increases numerically without limit as x  b, then n m

b

f(x)dx = limbn-

f(x)dx

m

a) If f(x) increases numerically without limit as x  c, a < c < b , (kumbaga yung point of discontinuity, hindi given pero nasa gitna siya ng a and b) then, b a

c

f(x)dx =

a

b

f(x)dx +

c

f(x)dx

n

= limnc-

b

f(x)dx + limmc+

a

f(x)dx

m

EXAMPLES: 2

1.

dx x(2 - x)

0

*pag sinubstite both 0 & 2, magiging undefined yung sagot so ii-integrate both limits b

lima0 and b2 b

dx 1 - (x - 2x + 1) 2

a

du 2 2 a -u

a

]

-1

b

Sin (x – 1) 𝑎 -1 -1 Sin (b – 1) – Sin (a – 1) -1 -1 Sin (1) – Sin (-1) π  π 2 ─ ─ 2  = π

* ─90° yung Sin-1(-1) instead of 180 kasi pag negative yung value tas Arcsin yung hinahanap, clockwise mo siya babasahin

EXERCERISES A. 1. A curve is such that y’’’ = 72x + 6 a. The curvature at any point (x,) on the curve: y’’= ______ b. The slop at any point (x,y) on the curve: y’ = _______ c. The general equation of the curve: y = _______ If the curve has a critical point at (0,1) and the curve also passes through (1,3): d. The values of the constants of integration: c1 = ____ c2 = ____ c3 = ____ e. At x = 1, y = _____ y’ = _____ 2. A stone that was tossed upward with a velocity of 16ft/sec from the top of a 96-ft high tower falls to the ground under 2 the influence of gravity only (g = 32ft /sec). Determine the equations of the motion of the stone as functions of time (Show the evaluation of the constants of integration): a. acceleration: a(t) = __________ b. velocity: v(t) = __________ c. displacement: s(t) = ___________ Based on the equations above, determine: d. time the stone takes before it hits the ground: t = ____s e. its velocity as it hits the ground: v = _____ft/s 2 3. Determine the area bounded by the curve y = x – 3x + 2 and the x-axis, from x = 0 to x = 2: a. A(x) = ___________________ b. the intersections of the curve with the x-axis: x1 = ____ x2 = ____ c. from x = 0 to x = x1 : c = ____ A = ____ d. from x = x1 to x = 2 : c = ____ A = ____ e. total area from x = 0 to x = 2: AT = ______ 2 dy (lnx) 4. Find the equation of the curve for which dx = x if the curve passes through (1,2).

B. Evaluate: 1.

2

2.

0

2.

dx 2/3 (x - 1) *If you substitute 0 & 2, the value will not be undefined. But if you substitute 1, it will be undefined. So you’ll apply the a < c < b rule.

1 0

dx 2/3 (x - 1) + b

limb13(x – 1)

2 1

]

1/3

1/3

4. 2

-2/3

(x – 2)

0

dx 2/3 (x - 1) dx + lima1+

b 0

+ 3(x – 1) 1/3

]

1/3

3.

2 𝑎

a

-2/3

(x – 2)

1/3

dx

1/3

3[ (b – 1) – (0 – 1) ] + 3[ (2 – 1) – (a – 1) 1/3 1/3 1/3 1/3 3[ (1 – 1) – (– 1) ] + 3[ (1) – (1 – 1) ] 3+3=6

5. ]

6. 7.

∫  ∫ ∫ ∫ ∫ ∫ ∫

1 x

2 x+

 2 dx 

(x - 2)(5x + 1) dx x 3

2

x + 2x + x dx 3

3x + 3x - 5 dx x-2 -2x

(x - e ) 2 -2x x + e dx 2 26

(x + x ) -1 (1 + 2x) dx 2

2 -1

(1 - 2x )(3 + lnx - x ) dx x

8.

∫

31.

∫

9.

∫ 3xex^2sin2ex^2 dx

32.

∫ x cosx2 (4sinx^2) dx

10.

∫

33.

xdx 4x^2 dx (3π)

 

16.

∫ ∫ ∫  ∫ ∫ ∫

17.

∫ (2x-3 + 3x2 + x-1)2 dx

11.

12. 13. 14.

(x

1/3

18. 19. 20. 21. 22.

23. 24. 25. 26.

2

-2

+ x -1)

dx

-1

(x + cosx) 2 cos (sinx + lnx) dx 1 2 2 2x ─ cos2x  csc (x – tanx) dx



2

2

sinh (x - cosx)(2x + sinx) 2 dx sech(x - cosx) 3

15.

2

2

x +x -3 x +5 dx x+1 2x 2 (3x + 2) dx

∫ ∫ ∫ ∫

e (-5x - 2) dx e

∫ ∫ ∫ ∫

cose -3x dx e

(2x + 1)

dx 2 x(1 + x ) x+6 2 (x + 2) dx

7 - lnx x(3 + lnx) dx 3 xdx dx 1 4x - 3 3x

cos3x 3 sin 3x dx cotx lnsinx dx 2

csc y coty 2 1 + csc y dy

29.

∫ (2x + 1) 4x2 + 4x - 3 dx ∫ sin2t 4 - cos2t dx ∫ x cotx2 cscx2 dx

30.

∫

27. 28.

34.

∫ ∫

35.

∫ (sinx + tanx)2 dx

x^2

(1 + e ) 2 x^2 x (x + e ) dx 1 x-2/3 + 2x x -1 3 -1

2t

e 2t 4t 1 + 6e + 9e dx

3

cos(tanx ) -2 2 3 x cos x dx

2/3

(1 + 6x ) 5/3 3 x + 6x dx (4 - tanx) 2 2 dx cos x 4 - tan x

36.

∫

38.

∫ p2(p3 +

dx dx x + 1 (x + 10) 3y y y 37. ∫ [ sin 2 + cos2 ] cos2 dx

39. 40. 41. 42.

∫ ∫

3

2.3 + ln4

5 )(p + 5 )

sinx 3 2 cos x(2 + tan x) dx 2 -1/2

(1 - 4x ) -4 (Arccos2x) dx

∫ sin32x(1 + cos4x) dx ∫ sin6x dx

dx

ANSWERS: A. 1. 2 a. 36x + 6x + c1 3 2 b. 12x + 3x + c1x + c2 2 4 3 c1x c. 3x + x + 2 + c2x + c3 d. c1 = -4 c2 = 0 c3 = 1 e. y = 3 y’ = 11 2. a. -32 b. -32t + 16 2

c. -16t + 16t d. 3 sec e. -80ft/s 3

2

x 3x 3. a. 3 ─ 2 + 2x + c b. x1 = 1

x2 = 2

c. c = 0

5 A=6

5 d. c = 6

1 A=6

e. AT = 1 s.u. 3

(lnx ) 4. y = 3 + 2 B. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

2

2x + 4x + lnx + c 5/2 3/2 1/2 2x – 6x – 4x + c 5/2 3/2 2x 2x 5 + 3 +c 3 2 x + 3x + 15x + 25ln(x – 2) + c 1 2 -2x 2 ln(x + e ) + c 2 27 (x + x ) 27 + c 2 ln(3 + lnx – x ) + c -4x^2 1 (3π) ─8 ln3π + c 3 x^2 ─4 cos2e + c 1 2 x^2 2 ln(x + e ) + c 1 1/3 2 3 3 (x + x - 1 ) + c tan(sinx + lnx) + x 2 ─cot(x – tanx) + c 3 2 sinh (x - cosx) +c 3 3 x 3 ─ 3x + 8ln(x + 1) + c

2 2 16. 9 [ ln(3x + 2) + 3x + 2 ] + c 5 4 9x 1 4 2 17. ─ 5x5 + 5 ─ x + 12lnx ─ 3x3 + 3x + c 1 7x + 3 18. 7 e +c 1 2 19. lnx ─ 2 ln(1 + x ) + c 4 20. ln(x + 2) ─ x + 2 21. 7ln(3 + lnx) ─ (3 + lnx) + 3ln(3 + lnx) + c 11 22. 6 1 3x 23. 3 sin(e ) + c 1 1 24. ─ 6 ▪ cos23x + c 25. ln(ln sinx) + c 1 2 26. 2 ln(1 + csc y) + c 1 2 3/2 27. 6 (4x + 4x – 3) + c 4 - cos2t 28. +c 3 1 2 29. ─2 cscx + c 2t 30. ln(1 + 3e ) + c 1 3 31. 3 sin(tanx ) sinx^2 4 32. 2ln4 + c 1/3 33. ln(x + 2x) + c -1tanx 2 34. 4 Sin 4 ─ 4 - tan x + c 1 1 35. 2 [x - 2 sin2x] + tanx – x + 2ln(secx + tanx) – 2sinx + c 2 -1 x + 1 36. 3 Tan 3 +c

1 1 y 1 cos2y + 2 4 2 + 2 siny + c (p3 + 5 )4.3ln4 12.9 + 3ln4 + c 1 2 2 ln(2 + tan x) + c 1 (Arccos2x)5 ─2 +c 5 3 5 cos 2x cos 2x 3 ─ 5 +c 1 3 5 1 sin32x 4 [ 2 x - 4 sin2x + 4 sin4x + 6 ]

37. ─ cosy – 38. 39. 40. 41. 42.