Integral Calculus Reviewer.pdf

INTEGRAL CALCULUS REVIEWER (2nd Sem 2011–2012)

Integration – process of a function whose derivative or

differential is given

Integrand – the given function Integral – the required function

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)

INDEFINITE INTEGRAL – if f(x) is a function whose derivative

is f(x), the relation between the two is given by:

∫

f(x) dx = F(x) + c Where: ∫ = integral sign f(x) = integrand F(x) = particular integrand c = constant of integration F(x) + c = indefinite integral of….

PROPERTIES: 1.

∫

du = u + c 2.

∫

(du + dv – dw) =

∫

du +

∫

dv –

∫

dw 3.

∫

Rdu =

∫

du c POWER FORMULAS:

∫

xndx = x n+1 n + 1 + c  if n ≠ –1

∫

x–1dx = lnx + c  if n = –1 EXAMPLES: 1.

∫

(3x2 – 6 x – _x94 ) dx

∫

3x2dx –

∫

6x1/2dx –

∫

9x–4dx 3

∫

x2dx – 6

∫

x1/2dx – 9

∫

x–4dx n = 2 n = 1₂ n = -4 3 ▪ x 3 3 – 6 ▪ x1/2 3 2 – 9 ▪ x -3 -3 + c X3 – 4x3/2 + 3x-3 + c 2.

∫

(x 2 - 3)(x2 + 2) 3 _x dx

∫

(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 n = 11_{3 n =} 5_{3 n =─} 1₃ 3x14/3 14 – 3x8/3 8 – 9x 2/3 + c 3.

∫

3 3x2 - 6 x x1/3 - 5 3 _2x3 dx

*put all x’s outside radicals

∫

3

3 x2/3 - 6x5/6 - 5

3 _{2 x} dx

*bring out constant denominator and place variable denominator in the numerator 1 3 ₂

∫

( 3 3 x2/3 – 6x5/6 – 5 ) ▪ x–1dx 1 3 2

∫

( 3 3 x2/3▪ x–1 – 6x5/6 ▪ x–1 – 5 ▪ x–1 ) dx 1 3 ₂

∫

( 3 3 x–1/3 – 6x–1/6 – 5 x–1) dx n = –1_{3 n = –}1_{6 n = –1} 1 3 ₂









3 3 ▪ 3x2/3 2 ─ 6 ▪ 6x5/6 5 ─ 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 y 2 ? dy = (2x-5)dx

∫

dy =

∫

(2x-5)dx y + c = x2 – 5x + c

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

1st option: y + c = x2 – 5x

*substitute x & y to get c

2 + c = (-1)2 – 5(-1) c = 4

*then substitute c & x to y + c = x2 – 5x

y + 4 = (4)2 – 5(4)

1st option: y = x2 – 5x + c

*substitute x & y to get c

2 = (-1)2 – 5(-1) + c c = -4

*then substitute c & x to y = x2 – 5x + c

y = (4)2 – 5(4) + (-4)

y = -8

2. Find the equation of the curve if the slope at pt (2,3) is given by 2x + 1_{2y - 3 .}

slope = dy_{dx =} 2x + 1_{2y - 3}  (2y – 3)dy = (2x + 1)dx

∫

(2y – 3)dy =

∫

(2x + 1)dx y2 – 3y = x2 + x + c *substitute pt (2,3) 32 – 3(3) = 22 + 2 + c c = -6 Equation: y2 – 3y = x2 + x – 6 (hyperbola)

3. If at any point (x,y) on a curve d 3

y

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.

d3y dx3 = 2 d dx









d2y dx2 = 2 d

_



d

_



2 y dx2 = 2dx

∫

d

_



d

_



2 y dx2 =

∫

2dx d2y dx2 = 2x + c1 d dx









dy dx = 2x + c1 d

_



dy_{dx = (2x + c}

_



1)dx

∫

d

_



dy_{dx =}

_



∫

(2x + c1)dx dy dx = x 2 + c1x + c2 dy = (x2 + c1x + c2)dx

∫

dy =

∫

(x2 + c1x + c2)dx y = x 3 3 + c1x2 2 + c2x + c3 SYSTEM OF EQUATIONS:

a. (1,3) is a point on a curve. So we substitute it to the last equation.

3 = 1_{3 +} c_{2 + c}1 2 + c3 16 = 3c1 + 6c2 + 6c3

b. slope = dy_{dx = -2 at x = 1. Substitute these} values to the second equation.

-2 = 12 + c1(1) + c2 c1 + c2 = -3 d2y dx2 = 2x + c1 0 = 2 + c1` c1 = -2

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

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

3

*substitute these values to the last equation

y = x 3 3 – x

2

– x + 14₃

4. Find the area under the parabola y = 8 – x2 – 2x, above the x-axis.

*complete the square

x2 + 2x + __ = 8 – y + __ x2 + 2x + 1 = 9 – y (x + 1)2 = –(y – 9)

*it is a parabola the opens downward V (-1,9) dA = (yA – yB)dx *dx = xLEFT – xRIGHT dA = (8 – x2 – 2x)dx

∫

dA =

∫

(8 – x2 – 2x)dx A = 8x – x 3 3 – x 2 _{+ c} On the x-axis, y = 0. *substitute y in y = 8 – x2 _{– 2x} 0 = 8 – x2 – 2x x2 + 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.

*Substitute these values to A = 8x – x

3 3 – x 2 + c 0 = 8(-4) – (-4) 3 3 – (-4) 2 _{+ c} c = 80₃ When x = 2, and c = 80₃

*Substitute these values to A = 8x – x

3

3 – x2 + c. We did this again tapos with x = 2 kasi may area nang

macocover sa point na yun . A = 8(2) – 2 3 3 – 2 2 + 80_{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 dt = 5t

2/3

+ 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 dt = 5t 2/3 _{+ 10t + 50}

∫

du =

∫

(5t2/3 + 10t + 50)dt u = 5t 5/3 5 3 + 5t2 + 50t + c u = 3t5/3 + 5t2 + 50t + c u 1000 ? t 0 4 When u = 1000 and t = 0. *Substitute to u = 3t5/3 _{+ 5t}2 _{+ 50t + c} c = 1000 When t = 4 and c = 1000 *Substitute to u = 3t5/3 + 5t2 + 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? a = g = dv_{dt = -32ft/s}2 *negative yung acceleration/gravity kasi opposite siya ng direction ng velocity ng hot air balloon

dv dt = -32

∫

dv =

∫

-32dt v = -32t + c1

When t = 0 and v = 10ft/s

*substitute these values to v = -32t + c1 10 = -32(0) + c1 c1= 10 v = ds_{dt = -32t + c}1

∫

ds =

∫

(-32t + c1)dt s = -16t2 + c1t + c2 When t = 0 and s = 0

*substitute these values to s = -16t2 + c1t + c2

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

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

s = -16t2 + 10t + 0 When s = -150

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

-150 = -16t2 + 10t 16t2 – 10t – 150 = 0

*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 

Differentiate s = -16t2 + 10t to get ds_{dt /the velocity} ds

dt = -32t + 10 When t = 3.4. v = -32(3.4) + 10

v = -98.8 ft/s

*Again, it is negative kasi opposite siya nung initial direction DEFINITE INTEGRAL PROPERTIES: 1.

b a f(x)dx = -

b a f(x)dx

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

b a f(x)dx =

c a f(x)dx +

b c f(x)dx

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

3.

b a f(x)dx =

c a f(t)dt +

b c f(z)dz

- The definite integral of an integrand is independent of the variable of integrations

EXAMPLES: 1.

2 -1





5x





2 + 1_{3 x –} 1_{2 dx} 5x3 3 + 1 6 x 2 – 1_{2 x}

]

2 −1 s = vot + at2 v = ds_dt a = g = dv_dt

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

5 3 (2 3 – (-1)3) + 1_{6 (2}2 – (-1)2) - 1_{2 (2+1)} 5 3 (8 + 1) + 1 6 (4 – 1) - 1 2 (2+1) = 28 2 = 14 2.

1 0 x3 + 1 x + 1 dx

₀1(x + 1)(x 2 _{- 2x + 1)} x + 1

1 0 (x 2 _{– 2x + 1)dx} x3 2 ─ x 2 _{+ x}

]

1 0 1 3 (1 3 _{– 0) – (1}2 _{– 0) + (1 – 0) =} 5 6

THE GENERAL POWER FORMULA

∫

undu  if n ≠ -1:

∫

undu = u n + 1 n + 1 + c  if n = -1:

∫

undu = lnu + c 1.

∫

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

∫

u2du u3 3 + c (x + 1)3 3 + c 1 3 (x 3 + 3x2 + 3x + 1) + c 1 3 x 3 + x2 + x + 1_{3 + c} 2.

∫

_{(2x - 7)}dx 4

∫

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

*trinanspose yung 2 sa other side para maging equal yung value ng du sa original formula. Pero yung 2 na trinanspose aka yung 1

2 , gagawin mong constant. so if you like, hide 1

2 , du = dx. So it still follows the original formula na ∫und. Pag hindi ‘to nagets explain ko sa other examples. =)))

1 2

∫

u -4 du 1 2 ▪ u-3 -3 + c ─_{6(2x - 7)}1 3 + c 3.

∫

tdt 4t2 + 9

∫

(4t2 + 9)-1/2tdt u = 4t2 + 9 du = 8tdt du 8 = tdt n = -1₂

* 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 un_{du. 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 8

∫

u -1/2 du 1 8 ▪ u1/2 1 2 + c 1 4 4t 2 + 9 + c 4.

∫

e 2t dt e4t + 22t + 1

∫

e2tdt (e2t + 1)2

∫

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

∫

u -2_du 1 2 ▪ u-1 -1 + c ─ _2e2t1 + 1 + c 5.

∫

y 1/3 (y4/3 + 9)2 dy

∫

(y4/3 + 9)-2y1/3dy u = y4/3 + 9 du = 4_{3 y}1/3dy 3 4 du = y 1/3 dy

3 4

∫

u -2_du 3 4 ▪ u-1 -1 + c ─ _4(y4/33 + 9) + c 6.

∫

(1 + 2e3x)e3xdx u = 1 + 2e3x du = 6e3xdx du 6 = e 3x dx 1 6

∫

u 1 du 1 6 ▪ u2 2 + c (1 + 2e3x)2 12 + c 7.

∫

x 3/4 _{+ 9} x1/4 dx n = 1₂ u = x3/4 + 9 du = 3_{4 x}-1/4dx 4 3 du = dx x1/4 4 3

∫

u 1/2_du 4 3 ▪ 2 3 u 2/3 _{+ c} 8 9 (x 3/4 _{+ 9)}2/3 _{+ c}

8.

∫

(6cos2x + sin2x)1/2sinxcosx dx

n = 1₂

u = 6cos2x + sin2x

du = 6[2cosx ▪ d(cosx)_{dx ] + 2cosx ▪} d(sinx)_dx du = 6 [2cosx(-sinx)]dx + 2sinxcosxdx du = -12sinxcosxdx + 2sinxcosxdx du = -10sinxcosxdx ─du_{10 = sinxcosxdx} ─₁₀ 1

∫

u1/2du ─_{10 ▪} 1 2_{3 u}3/2 + c ─_{15 (6cos}1 2x + sin2x)3/2 + c 9.

10 8









1 4 x - 1 -3 dx n = -3 u = 1_{4 x – 1} du = 1_{4 dx} 4du = dx 4 u-3du 4u-2 -2 ─ _u22 = ─ 2 (1₄ x - 1)2

]

10 8 ─ 2 (₄ 1(10) - 1)2 + 2 (1₄ (8) - 1)2 = 10 9 EXAMPLES

_











when

∫

du_{u or u}-1du = lnu + c : 1.

∫

sec5θtan5θ_{3 + 2sec5θ dθ} u = 3 + 2sec5θ du = 2(5)sec5θtan5θ dθ du 10 = sec5θtan5θ dθ 1 10

∫

du u 1 10 lnu + c 1 10 ln(3 + 2sec5θ) + c 2.

∫

dx x + x *factor x + x for it to be x ( x + 1)

∫

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

∫

du_{u = 2lnu + c} 2ln( x + 1) + c 3.

∫

secxdx

∫

secxtanx + sec_{secx + tanx}2x dx u = secx + tanx du = (secxtanx + sec2x)dx

∫

du u = ln(secx + tanx) + c 4.

ln2 0 exdx 1 + 3ex u = 1 + 3ex du = 3exdx du 3 = e x dx

*change the limits. To do that, substitute sa limits sa mga x sa equation ng u which is 1 + 3ex. when x = ln2, eln2 = 2. So 1 + 3(2) = 7. And when x = 0, it’s going to be e0 = 1. So 1 + 3(1) = 4.

x ln2 0 u 7 4 1 3

∫

du u 1 3 ln(1 + 3e x₎

]

7 4 1 3 [ln7 – ln4] 1 3 ln 7 4 5.

∫

x 3 - 2x + 5 x - 3 dx

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

∫

(x2 + 3x + 7 + _{x - 3 )dx} 26 x3 3 + 3x2 2 + 7x + 26

∫

(x – 3) -1_dx x3 3 + 3x2 2 + 7x + 26ln(x – 3) 6.

-2 -3 y + 2 y2 + 4y dy u = y2 + 4y du = (2y + 4)dy du 2 = (y + 2)dy *change limits x -3 -2 u -3 -4 1 2

-2 -3 du u = 1 2 ln |u|

]

−4 −3 1 2 [ ln|-4| - ln|-3|] 1 2 ln 4 3 EXPONENTIAL FUNCTION

∫

audu = _{lna a}1 u + c

∫

eudu = eu + c TRIGONOMETRIC FUNCTIONS 1.

∫

sin u du = – cos u + c 2.

∫

cos u du = sin u + c 3.

∫

tan u du = ln sec u + c = – ln cos u + c 4.

∫

cot u du = ln sin u + c = – ln csc u + c 5.

∫

sec u du = ln(sec u + tan u ) + c 6.

∫

csc u du = ln(csc u – cot u) + c 7.

∫

sec2u du = tan u + c

8.

∫

csc2u du = –cot u + c 9.

∫

sec u tan u du = sec u + c 10.

∫

csc u cot u du = –csc u + c EXAMPLES: 1.

∫

sin4xdx u = 4x du = 4dx du 4 = dx 1 4

∫

sin u du 1 4 (-cos u du) + c ─ 1_{4 cos4x + c} 2.

∫

tan x x dx u = x du = 1 2 x dx 2du = dx x 2

∫

tan u du = 2lnsec u + c 2ln(sec x ) + c 3.

∫

e2xcos e2x dx u = e2x du = 2e2x dx 2du = e2x _dx 1 2

∫

cos u du 1 2 sin e 2x + c

TRIGONOMETRIC TRANSFORMATIONS I.

∫

sinmx cosnx dx

where m or n is a positive odd integer tools: change the one w/ odd powers

sin2x = 1 – cos2x cos2x = 1 – sin2x Ex:

∫

sin52x cos42x dx y = 2x dy = 2dx dy 2 = dx 1 2

∫

sin 5_{y cos}4_{y dy} 1 2

∫

sin 4 y cos4y siny dy 1 2

∫

(sin 2 )2 cos4y siny dy 1 2

∫

(1 – cos 2 y)2 cos4y siny dy 1 2

∫

(1 – 2cos 2

y + cos4y) cos4y siny dy 1

2

∫

(cos 4

y – 2cos6y + cos8y) siny dy

*integrate each term. so their n’s sa un would be 4, 6, and 8 respectively.

u = cosy du = -siny dy

So we’ll be using theform ∫un = u

n+1

n + 1 for each term. And substitute 2x to y na ulit. 1 2

∫

(u 4 _{– 2u}6 _{+ u}8_{) siny dy} –1₂

_



cos

_



5_2x 5 – 2cos72x 7 – cos92x 9 + c –1_{2 cos}52x

_



1_{5 –} 2cos

_



2 2x 7 – cos42x 9 + c

II.

∫

secmx tannx dx or

∫

cscmx cotnx dx

a. Where m is positive even integer tools: sec2x = 1 + tan2x

csc2x = 1 + cot2x Ex:

∫

tan41_{2 x sec}4_{2 x dx} 1 y = 1_{2 x} 2dy = dx 2

∫

tan4y sec4y dy 2

∫

tan4y sec2y sec2y dy 2

∫

tan4y (1 + tanyx) sec2y dy 2

∫

(tan4y + tan6y) sec2y dy

u = tany du = sec2y 2

∫

(u4 + u6)du = 2

_











tan51₂ x 5 + tan71₂ x 7 + c

b. Where n is a positive odd integer tools: tan2x = sec2x – 1

cot2x = csc2x – 1 Ex:

∫

tan53x sec33x dx y = 3x dy 3 = dx 1 3

∫

tan 5 y sec3y dy 1 3

∫

tan 4

y sec2y tany secy dy 1

3

∫

(sec

2_{y – 1)}2 _sec2_{y tany secy dy} 1

3

∫

(sec

4_{y – 2sec}2_{y + 1) sec}2_{y tany secy dy} 1

3

∫

(sec

6_{y – 2sec}4_{y + sec}2_{y) tany secy dy}

u = secy du = tany secy dy 1 3

∫

(u 6 _{– 2u}4 _{+ u}2_)du 1 3









sec73x 7 – 2sec53x 5 + sec33x 3 + c

∫

_







1 + sinθ cosθ 2 dθ

∫

1 + 2sinθ + sin2θ cos2θ dθ

∫

1 cos2θ dθ + ∫ 2sinθ cos2θ dθ + ∫ sin2θ cos2θ dθ

∫

sec2θ dθ + 2

∫

secθ ▪ _{cosθ dθ +} sinθ

∫

tan2θ dθ 2tan2θ + 2

∫

secθ tanθ dθ +

∫

(sec2θ – 1) dθ 2tan2θ + 2secθ +

∫

sec2θ dθ –

∫

dθ

2tan2θ + 2secθ + tanθ – θ + c

III.

∫

tannx dx or cotnx dx

where n is an integer tools: tan2x = sec2x – 1

cot2x = csc2x – 1

a. n is a positive even integer

EX:

∫

tan6x dx

∫

tan4x ▪ tan2x dx

∫

(tan4x sec2x – tan4x) dx

*step by step nating i-solve each part, okay? So we’ll start with ∫tan4x sec2x

∫

tan4x sec2x dx u = tanx du = sec2x dx

∫

u4du u5 5 = tan5x 5 *next is –

∫

tan4x dx –

∫

tan4x dx –

∫

tan2x ▪ tan2x dx –

∫

tan2x(sec2x – 1) dx –

∫

(tan2x sec2x – tan2x) dx

*it’s possible na to integrate tan2x sec2x. Use

∫

undu. And distribute the negative sign so magiging positive yung tan2x. –tan 3 x 3 +

∫

tan 2 x dx –tan 3 x 3 +

∫

(sec 2 x – 1) dx –tan 3 x 3 +

∫

sec 2 x dx –

∫

dx –tan 3 x 3 + tan x – x

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

tan5x 5 –

tan3x

3 + tanx – x + c

b. n is a positive odd integer

EX:

∫

tan5x dx

∫

tan3x ▪ tan2x dx

∫

tan3x(sec2x – 1) dx

∫

tan3x sec2x dx –

∫

tan3x dx

*pwede na ma-integrate yung first term using undu so we’ll focus on the second term which is tan3xdx tan4x 4 –

∫

tan 2_{x ▪ tanx dx} tan4x 4 –

∫

tanx(sec 2_{x – 1) dx} tan4x 4 –

∫

(tanx sec 2_{x – tanx) dx} tan4x 4 –

∫

(tanx sec 2 x) dx –

∫

tanx dx tan4x 4 – tan2x 2 – ln(secx) + c

IV.

∫

sinmx cosnx dx

where m & n are positive even integers tools: sinx cosx = 1₂ sin2x

sin2x = 1₂ (1 – cos2x) cos2x = 1₂ (1 + cos2x) Ex:

∫

sin23x cos23x dx y = 3x dy = 3dx dy 3 = dx 1 3

∫

sin 2 y cos2y dx 1 3

∫

(siny cosy) 2 dx 1 3

∫

( 1 4 sin 2_2y)dy 1 12

∫

sin 2_2ydy 1 12

∫

1 2 (1 – cos4y)dy 1 24

∫

dy – 1 24

∫

cos4ydy 1 24 [y – 1 4 sin4y] + c 1 24 [3x – 1 4 sin12x] + c

∫

sin2x cos4x dx

∫

sin2x cos2x cos2x dx

∫

(sinx cosx)2 cos2x dx

∫

(1_{2 sin2x)}2cos2x dx 1 4

∫

sin 2_{2x cos}2_{x dx} 1 8

∫

(1 – cos4x) cos 2_{x dx} 1 8

∫

cos 2 x dx – 1₈

∫

cos4x cos2x dx 1 8

∫

1 2 (1 + cos2x) dx – 1 8

∫

cos4x ▪ 1 2 (1 + cos2x) dx 1 16

∫

(1 + cos2x) dx – 1

16

∫

(cos4x + cos2x cos4x) dx

1 16 (x + 1 2 sin2x) – 1 16

∫

cos4x dx – 1 16

∫

(cos2x cos4x) dx 1 16 (x + 1 2 sin2x) – 1 16 ▪ 1 4 sin4x – 1 16

∫

(cos2x(1 – 2sin 2 2x) dx 1 16 (x + 1 2 sin2x) – 1 64 sin4x – 1 16

∫

(cos2x – 2sin 2_{2x cos2x) dx} 1 16 (x + 1 2 sin2x) – 1 64 sin4x – 1 16

∫

cos2x dx – 1 16 ▪ 2

∫

sin 2_2x

1 16 (x + 1 2 sin2x) – 1 64 sin4x – 1 16 ▪ 1 2 sin2x – 1 8 ▪ 1 2

∫

u 2_du 1 16 x + 1 32 sin2x – 1 64 sin4x – 1 32 sin2x – 1 16 ▪ u3 3 + c 1 16 x + 1 32 sin2x – 1 64 sin4x – 1 32 sin2x – 1 16 ▪ u3 3 + c 1 16 x + 1 32 sin2x – 1 64 sin4x – 1 32 sin2x – 1 48 sin 3 2x + c

V.

∫

sin ax sin bx dx

∫

sinmx cosnx dx

∫

sinmx cosnx dx

tools: sinα sinβ = 1_{2 *cos(α ─ β) – cos(α + β)]}

cosα cosβ = 1_{2 *cos(α ─ β) + cos(α + β)+}

sinα cosβ = 1_{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 2 [ 1 3 sin3x – 1 11 sin11x] + c 2.

∫

cos7x sin4x dx *let α = 4x and β = 7x 1 2

∫

[sin(4x – 7x) + sin(4x + 7x)] dx 1 2

∫

[sin(–3x) + sin(11x)] dx 1 2

∫

(–sin3x + sin11x) dx 1 2 [ 1 3 cos3x – 1 11 cos11x] + c 3.

π 4 0 cosx cos3x dx 1 2

∫

[cos(x – 3x) + cos(x + 3x)] dx 1 2

∫

(cos2x + cos4x) dx 1 2

[

1 2 sin2x + 1 4 sin4x

]

π 4 0 1 2

[









1 2 sin2( π 4 ) + 1 4 sin4( π 4 ) –









1 2 sin2(0) + 1 4 sin4(0)

]

1 2

[

1 2 (1) + 1 4 (0)

]

– 0 = 1 4 4.

π 3

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.

1 2 ▪ sin2x 2 – 1 2

∫

1 2 [sin(x – 5x) + sin(x + 5x)] dx sin2x 4 – 1 4

∫

(-sin4x + sin6x) dx sin2x 4 + 1 4

∫

sin4x dx – 1 4

∫

sin6x dx sin2x 4 + 1 16 cos4x – 1 24 cos6x

]

π 3 0 = 9 32

INVERSE TRIGONOMETRIC FUNCTIONS

1.

∫

du a2 ─ u2 = Sin -1u a + c 2.

∫

_a2du _{+ u}2 = 1_{a Tan}-1u_{a + c} 3.

∫

du u u2 ─ a2 = 1 a Sec -1 u a + c Examples: 1.

∫

_{25 + 64x}dx 2

∫

dx (5)2 + (8x)2 a = 5 u = 8x du 8 = dx 1 8

∫

du a2 + u2 1 8 ▪ 1 5 Tan -18x 5 + c 1 40 Tan -18x 5 + c

2.

∫

dx 9 ─ 4x2

∫

dx (3)2 ─ (2x)2 a = 3 u =2x du 2 = dx 1 2

∫

du a2 ─ u2 1 2 Sin -12x 3 + c 3.

∫

sec 2_{x dx} 50 ─ sec2x

∫

sec2x dx 50 ─ (1 + tan2x)

∫

sec2x dx 49 ─ tan2x

∫

sec2x dx

(7)2 ─ (tanx)2 a = 7 u =tanx du = sec

2

x dx

Sin-1

_



tanx_{7 + c}

_



4.

∫

dx

21 - 4x + x2

*add and subtract 4 para maging perfect square yung x2 _{– 4x}

∫

dx

21 + 4 ─ (x2 ─ 4x + 4)

*nakalagay sa equation, + 4 sa pareho, kasi yung second na 4, negative siya pag dinitribute yung nega

∫

dx

25 ─ (x ─ 2)2 a = 5 u = x – 2 du = dx

Sin-1

_



x ─ 2_{5 + c}

_



5.

∫

dx

5 - 2x - 3x2

*to get c in ax2 _{+ bx + c, get the value of} b 2 4a . so in this formula c = 2 2 4(3) = 1 3

∫

dx 5 + 1_{3 ─ (3x}2 + 2x + 1_{3 )}

∫

dx









4 3 2 _{─ ( 3 x +} 1 3 ) 2 a = 4 3 u = 3 x + 1 3

∫

du a2 ─ u2 du 3 = dx 1 3 Sin -1









3x + 1 4 + c 6.

∫

_5x2 dx - 4x + 2

∫

dx (5x2 - 4x + 4_{5 ) + (2 -} 4_{5 )}

∫

dx ( 5 x ─ 2 5 ) 2 +

_



6

_



5 2 a = 2 5 u = 5 x ─ 2 5

∫

du a2 + u2 du 5 = dx 1 5 ▪ 5 6 Tan -1













5x - 2 5 6 5 + c 1 6 Tan -15x - 2 6 + c 7.

∫

dx x 9x2 - 25

*multiply the whole equation to 3_{3 para maging 3x} yung x na nasa baba.

∫

3dx 3x (3x)2 - (5)2 a = 5 u = 3x du 3 = dx

∫

du u u2 ─ a2 1 5 Sec -13x 5 + c 8.

1 0 (x + 1)dx x2 + 1

∫

x x2 + 1 dx +

∫

dx x1 + 12 *For _x2x + 1 : u = x2 + 1 du_{2 = xdx}

∫

duu + ∫ dx x1 + 12 *For _x1dx + 12 : u = x a = 1 du = dx 1 2

∫

du u +

∫

du u2 + a2

1 2 ln(x 2 + 1) + Tan-1x

]

1 0 1 2 [(ln1 2 + 1) – (ln02 +1)] + [Tan-11 – Tan-10] 1 2 [ln2 – ln1] + [ π 4 - 0] 1 2 ln2 + π 4 9.

e 1 dy y(1 + ln2y)

∫

dy

y(12 + (lny)2) a = 1 u = lny du =

dy y

∫

du a2 + u2 Tan-1(lny)

]

e 1

Tan-1(lne) – Tan-1(ln1) Tan-11 – Tan-10 π 4 10.

4 2 dy y y ─ 1

∫

dy y ( y )2 ─ 12

*Diba u = y . So yung y sa labas, ihiwalay mo para

maging 1

y

y . Para yung form maging du u u2 ─ a2 u = y 2du = dy y 2

∫

du u u2 ─ a2 2 Sec-1 y

]

4 2 2 [Sec-1 4 ─ Sec-1 2 ] 2 [ π_{3 ─} π_{4 ] = 12} π 11.

6 2 (5x - 2)dx x2 - 4x + 20 *let u = x2 – 4x + 2z and du = (2x – 4)dx *divide the numerator by the derivative of the denominator. Then follow this:

∫

52 (2x - 4) + 8 x2 - 4x + 20 5 2∫ (2x - 4)dx x2 - 4x + 200 + 8∫ dx x2 - 4x + 200 u = x2 _{– 4x + 200} du = 2x – 4 5 2

∫

du u + 8

∫

dx (x2 - 4x + 4) + (20 - 4) 5 2 ln(x 2 _{– 4x + 20) + 8}

∫

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

∫

du a2 + u2 5 2 ln(x 2 – 4x + 20) + 8 ▪ 1_{4 Tan}-1

_



x - 2₄

_



]

6 2 5 2 [ln32 – ln16] + 2 [Tan -1 1 – Tan-10] 5 2 ln 32 16 + 2 ▪ π 4 5 2 ln2 + π 2 12.

2 1 (3x + 1)dx 3 + 2x - x2 u = 3 + 2x – x 2 _{du = (2 – 2x)dx}

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

∫

─3_{2 (-2x + 2) + 4} 3 + 2x - x2 dx ─32 ∫ -2x + 2 3 + 2x - x2 dx + 4

∫

dx 3 + 2x - x2 u = 3 + 2x – x2 _{du = (2 – 2x)dx} ─32 ∫u-1/2du + 4

∫

dx (3 + 1) - (x2 - 2x + 1) ─3_{2 ▪} 2u 1/2 1 + 4

∫

dx (2)2 - (x- 1)2 – 3 3 + 2x - x2 + 4

∫

du a2 ─ u2 – 3 3 + 2x - x2 + 4 Sin-1

_



x - 1₂

_



]

2 1 *─3 3 – (-6)] + 4[ π_{6 ─ 0+} 6 - 3 3 + 2π₃ 13.

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

∫

dx

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

∫

dx (x + 1) x2 + 2x + 1 - 1 1 2

∫

dx (x + 1) (x + 1)2 - 12 u = x + 1 du = dx a = 1

1 2

∫

du u u2 ─ a2 1 2 Sec -1 (x + 1)

]

2 1 1 2 [Sec -1_{3 – Sec}-1_2] 1 2 [Sec -1 3 – π_{3 ]} ADDITIONAL FORMULAS: 1.

∫

u2 ± a2 du = 1_{2 { u u}2±a2 ± a2 ln

|

u + u2±a2

|

} + c 2.

∫

du u2 ± a2 = ln

|

u + u 2_±a2

_|

_{} + c} 3.

∫

a2 ─ u2 du = 1_{2 { u a}2 ─ u2 + a2 Sin-1

_



u_{a } + c}

_



4.

∫

_u2du - a2 = 1 2a ln

|

u - a u + a

|

+ c 5.

∫

_a2du - u2 = 1 2a ln

|

u + a u - a

|

+ c EXAMPLES: 1.

∫

xdx 9x4 - 1

∫

xdx (3x2)2 - 12 u = 3x 2 du_{6 = xdx a = 1} 1 6

∫

du u2 ± a2 1 6 ln

|

3x 2 + 9x4 - 1

|

+ c 2.

∫

dx x2 + x + 1

∫

dx x2 + x + 1₄ + (1 - 1₄ )

∫

dx (x + 1_{2 )}2 + ( _{2 )}3 2 a = x + 1_{2 du = dx a = 2} 3

∫

du u2 ± a2 ln

|

(x + 1_{2 + x}2 + x + 1

|

+ c 3.

∫

_x2 dx - 3x - 10

∫

dx (x2 - 3x + 9₄ ) - (10 + 9₄ )

∫

dx (x- 3₂ )2 - 49₄ u = x - 3 2 du = dx a = 7 2

∫

du u2 - a2 1 7 ln

|

(x - 3₂ ) - 7₂ (x - 3₂ ) + 7₂

|

+ c 1 7 ln

|

x - 5 x + 2

|

+ c 4.

1 0 dx 2 - x2

∫

dx ( 2 )2 - x2 u = x du = dx a = 2

∫

du a2 - u2 1 2 2 ln

|

x + 2 x - 2

|

]

1 0 1 2 2 [ ln

|

1 + 2 1 - 2

|

─ ln

|

0+ 2 0 - 2

|

] 1 2 2 [ ln

|

1 + 2 1 - 2

|

─

|

ln1

|

] 1 2 2 ln

|

1 + 2 1 - 2

|

5.

4 3 25 - x 2 dx a = 5 u = x du = dx

∫

a2 ─ u2 du 1 2 { x 25 - x 2 _{+ 25 Sin}-1









x 5 }

]

4 3 1 2 { [12 + 25 Sin -1









4 5 ] – [12 + 25 Sin -1









3 5 25 2 [ Sin -1









4 5 – Sin -1









3 5 ]

*diba yung notation na “Sin-1_{(x)” means ANGLE yung}

value niya? Like, “Sin-1(1)” = 90 or π_{2 . So let’s represent} Sin-1

( )

4 5 = θ and Sin -1

( )

3 5 = β. 25 2 (θ – β)

*recall the identity sin(θ – β) = sinθcosβ – cosθsinβ. Draw ka ng triangle of each angle. Since sin function yung 4_{5 at} 3

5 , ile-label mo yung numbers na yan sa opposite hypotenuse . So the triangles would look like

sin(θ – β) = sinθcosβ – cosθsinβ sin(θ – β) =



_

4₅

_



_



4_{5 –}



_

_



3₅

_



_



3₅

_



sin(θ – β) = ₂₅ 7 *note that we’re only getting (θ – β)

(θ – β) = Sin-1

_



₂₅ 7

_



25 2 Sin -1









7 25 6.

π 4 π 12 cos2x sin22x - ₁₆1 dx a = 1 4 u = sin2x du 2 = cos2xdx 1 2

∫

du u2 - a2 1 2 ▪ 1 2(1₄ ) ln

|

sin2x - 1₄ sin2x + 1₄

|

]

π 4 π 12 ln

|

sin2

_



π_{4 -}

_



₄1 sin2

_



π_{4 +}

_



1₄

|

─ ln

|

sin2

_



_{12 -} π

_



1₄ sin2

_



_{12 +} π

_



1₄

|

ln

|

sin

_



π_{2 -}

_



1₄ sin

_



π_{2 +}

_



1₄

|

─ ln

|

sin

_



π_{6 -}

_



1₄ sin

_



π_{6 +}

_



1₄

|

ln

















3 4 5 4 ─

















1 4 3 4 ln

_



3_{5 ─ ln}

_



_



1₃

_



ln9₅ 7.

∫

_9y6y + 12 - 6y - 3 dy u = 9y2 – 6y – 3 du = (18y – 6)dy

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

∫

13 (18y - 6) + 3 9y2 - 6y - 3 dy 1 3

∫

(18y - 6)dy 9y2 - 6y - 3 + 3

∫

dy (9y2 - 6y + 1) - (3 + 1) 1 3

∫

du u + 3

∫

du u2 - a2 u = 3y + 1 du 3 = dy a = 2 1 3 ln(9y 2 _{– 6y – 3) +} 3 4 ln

|

3y - 3 3y + 1

|

+ c 8.

∫

2x - 3 x2 + x + 2 dx u = x 2 _{+ x + 2 du = (2x + 1)dx}

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

∫

(2x - 1) - 4 x2 + x + 2 dx

∫

(2x - 1) x2 + x + 2 dx ─ 4

∫

dx (x2 + x + 1₄ ) + (2 - 1_{4 )}

∫

u-1/2du ─ 4

∫

du u2 - a2 2(x2 + x + 2)1/2 – 4 ln

|

(x + 1₂ ) + x2 + x + 2

|

} + c 2 x2 + x + 2 ─ 4 ln

|

(x + 1₂ ) + x2 + x + 2

|

} + c 9.

e^3 e^2 lnx x(ln4x - 1) dx u = ln2x du 2 = lnx x dx a = 1 1 2

∫

du u2 - a2 1 2 ▪ 1 2 ln

|

ln2x - 1 ln2x + 1

|

]

e^3 𝑒^2 1 4









ln

|

ln 2_e3 _{- 1} ln2e3 + 1

|

─ ln

|

ln2e2 - 1 ln2e2+ 1

|

1 4 [ ln

|

33 - 1 33 + 1

|

─ ln

|

22 - 1 22 + 1

|

] 1 4 [ ln 4 5 ─ ln 3 5 ] = 1 4 ln 4 3 10.

∫

1 + 1_{x dx}

∫

x + 1 ( x )2 dx

∫

1 x ( x ) 2 + 12 dx u = x 2du = dx x a = 1 2

∫

u2 ± a2 du 1 2 { x ▪ x + 1 + ln

|

x + x + 1

|

} + c HYPERBOLIC FUNCTIONS 1.

∫

sinh u du = cosh u + c 2.

∫

cosh u du = sinh u + c 3.

∫

tanh u du = ln

|

cosh u

|

+c 4.

∫

coth u du = ln

|

sinh u

|

+c 5.

∫

sech2 u du = tanh u + c 6.

∫

csch2u du = –coth u + c

7.

∫

sech u tanh u du = –sech u + c

EXAMPLE:

∫

(sech 1 - t )(tanh 1 - t ) 1 - t dx u = 1 - t ─2du = dt 1 - t ─2

∫

sech u ▪ tanh u ▪ du ─2(–sech u) + c 2sech 1 - t + c IMPROPER INTEGRALS

I. Integrals with infinite limits in the integrand

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

_a∞f(x)dx = lim_b∞

b a f(x)dx

_-∞b f(x)dx = lim_a

_-

_∞

b a f(x)dx

_-∞∞ f(x)dx = lim_a

_-

_∞ and_b∞

b a f(x)dx 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 EXAMPLES: 1.

∞ 1 2dy y(y + 16) lim_b∞

b 1 2dy y2 + 16y + 64 - 64 2

b 1 dy (y + 8)2 - (8)2 a = 8 u = y + 8 du = dy

∫

du u2 - a2 2 ▪ _{2(8) ln} 1

|

_{y + 8 + 8} y + 8 - 8

|

2 ▪ _{2(8) ln} 1

|

_{y + 16} y

|

]

b 1 1 8









ln

|

_{b + 16} b

|

─ ln

|

_{1 + 16} 1

|

1 8









ln

|

∞_∞

|

─ ln

|

₁₇ 1

|

*so diba infinity over infinity, so bawal yun. Babalik tayo sa equation before this. Yung may b over b + 16. Derive that. 1 8









ln

|

1₁

|

─ ln

|

₁₇ 1

|

*recall that ln1 = 0 ─ 1_{8 ln 17} 1

*recall that lna_{b = lna - lnb}

─ 1_{8 [ln1 – ln17]} ─ 1_{8 [–ln17] =} 1_{8 [ln17]} 2.

∞ 0 xe -x^2 dx lim_b∞

b 1 xe -x^2 dx u = ─x2 _─ du 2 = xdx ─1₂

b 1 e u du ─1₂

_



_e1x^2

_



]

b 0 ─1₂

_



_e1b^2 ─ 1

_



e0 ─1₂

_



_{∞ ─} 1 1₁

_



─1_{2 ▪ ─1 =} 1₂ 3.

∞ 2 dx x2 + 1 lim_b∞

b 1 du u2 + a2 1 2a ln

|

u - a u + a

|

= 1 2 ln

|

x - 1 x + 1

|

]

b 2 1 2









ln

|

_{b + 1} b - 1

|

─ ln

|

_{2 + 1} 2 - 1

|

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

1 2









ln

|

1₁

|

─ ln

|

1₃

|

1 2









─ ln

|

1₃

|

─ 1_{2 ln} 1_{3 = ─} 1_{2 [ln1 – ln3] =} 1_{2 ln3}

II. Integrals with infinite discontinuities in the integrand

*in other words, isa or both a and b sa formula na _abf(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 f(x)dx = limam+

n a f(x)dx

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

_mn f(x)dx = lim_bn-

b m f(x)dx

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,

_ab f(x)dx =

c a f(x)dx +

b c f(x)dx = lim_nc-

n a f(x)dx + limmc+

b m f(x)dx EXAMPLES: 1.

2 0 dx x(2 - x)

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

lim_{a0 and b2}

b a dx 1 - (x2 - 2x + 1)

_ab _adu2 _{- u}2 Sin-1(x – 1)

]

b 𝑎 Sin-1(b – 1) – Sin-1(a – 1) Sin-1(1) – Sin-1(-1) π 2 ─





─





π 2 = π

* ─90° yung Sin-1_{(-1) instead of 180 kasi pag negative}

yung value tas Arcsin yung hinahanap, clockwise mo siya babasahin 2.

2 0 dx (x - 1)2/3

*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 (x - 1)2/3 +

2 1 dx (x - 1)2/3 lim_b1-

b 0 (x – 2) -2/3 dx + lim_a1+

2 a (x – 2) -2/3 dx 3(x – 1)1/3

]

b 0 + 3(x – 1) 1/3

]

2 𝑎 3[ (b – 1)1/3 – (0 – 1)1/3 ] + 3[ (2 – 1)1/3 – (a – 1)1/3 ] 3[ (1 – 1)1/3 – (– 1)1/3 ] + 3[ (1)1/3 – (1 – 1)1/3 ] 3 + 3 = 6 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 the influence of gravity only (g = 32ft2/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

3. Determine the area bounded by the curve y = x2 – 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 = ______ 4. Find the equation of the curve for which dy_{dx =} (lnx)

2 x if the curve passes through (1,2).

B. Evaluate: 1.

∫

_



2 x + 1

_



x 2 dx 2.

∫

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

∫

x3 + 2x2 + x dx 4.

∫

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

∫

(x - e -2x₎ x2 + e-2x dx 6.

∫

(x + x 2₎26 (1 + 2x)-1 dx 7.

∫

(1 - 2x 2 )(3 + lnx - x2)-1 x dx

8.

∫

_(3π)xdx4x^2 dx 9.

∫

3xex^2sin2ex^2 dx 10.

∫

(1 + e x^2₎ x-1(x2 + ex^2) dx 11.

∫









1 3 x -2/3 + x x2 - 1 (x1/3 + x2 - 1 )-2 dx 12.

∫

(x -1 + cosx) cos2(sinx + lnx) dx 13.

∫

_



2x ─ _cos12_{x csc}

_



2(x2 – tanx) dx 14.

∫

sinh 2_(x2 _{- cosx)(2x + sinx)} sech(x2 - cosx) dx 15.

∫

x 3 + x2 - 3 x2 + 5 x + 1 dx 16.

∫

_{(3x + 2)}2x 2 dx 17.

∫

(2x-3 + 3x2 + x-1)2 dx 18.

∫

e (2x + 1) e(-5x - 2) dx 19.

∫

_{x(1 + x}dx 2₎ 20.

∫

_{(x + 2)}x + 62 dx 21.

∫

_{x(3 + lnx) dx} 7 - lnx 22.

3 1 xdx 4x - 3 dx 23.

∫

cose 3x e-3x dx 24.

∫

_sincos3x3 3x dx 25.

∫

_{lnsinx dx} cotx 26.

∫

csc 2 y coty 1 + csc2y dy 27.

∫

(2x + 1) 4x2 + 4x - 3 dx 28.

∫

sin2t 4 - cos2t dx 29.

∫

x cotx2 cscx2 dx 30.

∫

e 2t 1 + 6e2t + 9e4t dx 31.

∫

cos(tanx 3 ) x-2 cos2x3 dx 32.

∫

x cosx2 (4sinx^2) dx 33.

∫

(1 + 6x 2/3₎ 3 x + 6x5/3 dx 34.

∫

(4 - tanx) cos2x 4 - tan2x dx 35.

∫

(sinx + tanx)2 dx 36.

∫

dx x + 1 (x + 10) dx 37.

∫

[ sin3y_{2 + cos}y_{2 ] cos}y_{2 dx} 38.

∫

p2(p3 + 5 )(p3 + 5 )2.3 + ln4 dx 39.

∫

_cos3 sinx x(2 + tan2x) dx 40.

∫

(1 - 4x 2₎-1/2 (Arccos2x)-4 dx 41.

∫

sin32x(1 + cos4x) dx 42.

∫

sin6x dx

ANSWERS: A. 1. a. 36x2 + 6x + c1 b. 12x3 + 3x2 + c1x + c2 c. 3x4 + x3 + c1x 2 2 + c2x + c3 d. c1 = -4 c2 = 0 c3 = 1 e. y = 3 y’ = 11 2. a. -32 b. -32t + 16 c. -16t2 + 16t d. 3 sec e. -80ft/s 3. a. x 3 3 ─ 3x2 2 + 2x + c b. x1 = 1 x2 = 2 c. c = 0 A = 5₆ d. c = ₆ 5 A = 1₆ e. AT = 1 s.u. 4. y = (lnx 3 ) 3 + 2 B. 1. 2x2 + 4x + lnx + c 2. 2x5/2 – 6x3/2 – 4x1/2 + c 3. 2x 5/2 5 + 2x3/2 3 + c 4. x3 + 3x2 + 15x + 25ln(x – 2) + c 5. 1_{2 ln(x}2 + e-2x) + c 6. (x + x 2 )27 27 + c 7. ln(3 + lnx – x2) + c 8. ─1₈ (3π) -4x^2 ln3π + c 9. ─3_{4 cos2e}x^2 + c 10. 1_{2 ln(x}2 + ex^2) + c 11. 1_{3 (x}1/3 + x2 - 1 )3 + c 12. tan(sinx + lnx) + x 13. ─cot(x2 – tanx) + c 14. sinh 3_(x2 _{- cosx)} 3 + c 15. x 3 3 ─ 3x + 8ln(x + 1) + c 16. 2_{9 [ ln(3x + 2) + 3x + 2 ] + c} 2 17. ─ _5x45 + 9x 5 5 ─ 1 x + 12lnx ─ 4 3x3 + 3x 2 _{+ c} 18. 1_{7 e}7x + 3 + c 19. lnx ─ 1_{2 ln(1 + x}2) + c 20. ln(x + 2) ─ _{x + 2} 4 21. 7ln(3 + lnx) ─ (3 + lnx) + 3ln(3 + lnx) + c 22. 11₆ 23. 1_{3 sin(e}3x) + c 24. ─ 1_{6 ▪ cos}12 3x + c 25. ln(ln sinx) + c 26. 1_{2 ln(1 + csc}2y) + c 27. 1_{6 (4x}2 + 4x – 3)3/2 + c 28. 4 - cos2t ₃ + c 29. ─1_{2 cscx}2 + c 30. ln(1 + 3e2t) + c 31. 1_{3 sin(tanx}3) 32. 4 sinx^2 2ln4 + c 33. ln(x1/3 + 2x) + c

34. 4 Sin-1tanx_{4 ─ 4 - tan}2x + c

35. 1_{2 [x -} 1_{2 sin2x] + tanx – x + 2ln(secx + tanx) – 2sinx + c} 36. 2_{3 Tan}-1 x + 1 ₃ + c

37. ─

1

_{2 cosy –}

1

_{4 cos2y +}

y

_{2 +}

1

_{2 siny + c}

38.

(p

3

+ 5 )

4.3ln4

12.9 + 3ln4 + c

39.

1

_{2 ln(2 + tan}

2

x) + c

40.

─

1

₂

(Arccos2x)

5

5

+ c

41.

cos

3

2x

3

─

cos

5

2x

5

+ c

42.

1

_{4 [}

3

_{2 x -}

5

_{4 sin2x +}

1

_{4 sin4x +}

sin

3

2x

6 ]