Exam 2

Question 1

∫ 2+x^2 / 1+x^2 dx

Answer 1

Split into ∫ 2 / 1+x^2 dx + ∫ x^2 / 1+x^2 dx
Divide x^2+1 into x^2
Create new integral
∫ 2 / 1+x^2 dx + ∫ (1 - 1/x^2+1) dx
∫ 2 / 1+x^2 dx + ∫dx + ∫ 1 / 1+x^2
∫ 1 / 1+x^2 dx + ∫dx
= tan^-1(x) + x +C

Question 2

∫ tan^2ϴ * sec^4ϴ dϴ

Answer 2

Split sec^4ϴ into sec^2ϴ*sec^2ϴ
sec^2ϴ = (1+tan^2ϴ)
∫ tan^2ϴ(1+tan^2ϴ)sec^2ϴ dϴ
Distribute
u=tanϴ
du=sec^2ϴ dϴ
∫ u^2 du + ∫ u^4 du
=u^3/3 + u^5/5 + C
= tan^3ϴ/3 + tan^5ϴ/5 +C

Question 3

Integration by parts formula

Answer 3

∫ udv = uv - ∫vdu

Question 4

∫ lnx dx

Answer 4

Integration by Parts
u = lnx , du = 1/x dx
v = x , dv = dx
Sub into IBP
xlnx - ∫x * 1/x dx
xlnx - ∫ dx
= xlnx -x +C

Question 5

∫ ln (2x +1) dx

Answer 5

Integration by Parts
W sub ; w = 2x+1 , dw = 2dx , dx = dw/2
∫ lnw * dw/2
1/2 ∫ lnw dw
1/2 (wlnw - w +C) - IBP
Sub w back in
=1/2(2x+1) * ln(2x+1) - 1/2(2x+1) +C

Question 6

∫ tan^2x sec^2x dx

Answer 6

Integration by Parts
u = tan^2x , du = 2tanx*sec^2x dx
v = tanx , dv = sec^2x dx
tan^3x - ∫ tanx * 2tanxsec^2x dx
tan^3x - 2 ∫ tan^2xsec^2x dx
Original Integral I = tan^2xsec^2x dx
= tan^3x -2I
3I = tan^3x
I = tan^3x/3 +C

Question 7

∫ y / e^2y dy : [0,1]

Answer 7

Integration by Parts
w = -2y , y = w/-2
dw = -2dy
dy = -dw/2
∫ w/-2 * e^w * dw/-2
I = 1/4 ∫ we^w dw
J = ∫ we^w dw
u=w , dv = e^w dw
du = dw , v = e^w
= we^w - ∫ e^w dw = we^w -e^w +C
= -2ye^-2y - e^-2y +C
I = 1/4 J
=1/4 (-2ye^-2y - e^-2y +C)
=-1/2 * ye^-2y ] from 0,1 - 1/4 * e^-2y ] from 0,1

Question 8

∫ tany dy

Answer 8

Trig Integral
∫ siny/cosy dy
u = cosy , du = -siny dy
-du = siny dy
= - ∫ du/u
= -ln u +C
= -ln |cosy|+C
=ln |cosy|^-1 +C
=ln |secy| +C

Question 9

∫ cos^2xdx

Answer 9

Trig Integral
∫ 1 + cos2x / 2 dx
∫ 1/2 dx + ∫ cos2x/2 dx
=1/2x + 1/4sin2x +C

Question 10

∫ sin^2xdx

Answer 10

Trig Integral
∫ 1 - cos2x / 2 dx
∫ 1/2 dx - ∫ cos2x/2 dx
=1/2x - 1/4sin2x +C

Question 11

∫ dx / 1 + cosx

Answer 11

Trig Integral
∫ dx / 1 +cosx * (1-cosx/1-cosx)
∫ (1-cosx) / sin^2x dx
∫ 1/sin^2x dx - ∫ cosx / sin^2x dx
∫ csc^2x dx - ∫ cotx*cscx dx
-cotx - (-cscx) +C
= -cotx + cscx +C

Question 12

∫ xtan^-1x dx

Answer 12

Trig Integral
u = tan^-1x , x = tanu
du = 1 / 1+x^2 dx
dx = (1+x^2)du = (1 + tan^2u) du
I = ∫ tanu * u(1+tan^2u) du
=∫ utan^3u du + ∫ utanu du
First part cancels
= ∫ xtan^-1x dx
u = tan ^-1 , du = 1/(1+x^2) dx
v = x^2 / 2 , dv = xdx
= x^2 / 2 tan^-1x - ∫ x^2 / 2 * 1/(1+x^2) dx
Divide x^2 +1 into x^2
=> 1/2 ∫ (1- 1(x^2+1)) dx => 1/2 tan^-1x
I = x^2 / 2 tan^-1x - 1/2x + 1/2tan^-1x +C
= 1/2 (x^2+1) tan^-1x - 1/2x + C

Question 13

sqrt (a^2 - x^2)

Answer 13

Trig Sub
x = asin ϴ or x = acosϴ
= sqrt (a^2 - a^2sin^2ϴ)
= sqrt (a^2 (1-sin^2 ϴ)
= a * sqrt (cos^2ϴ)
=acosϴ

Question 14

sqrt (a^2 + x^2)

Answer 14

Trig Sub
x = atanϴ
= sqrt (a^2 + a^2tan^2ϴ)
= sqrt (a^2(1+tan^2ϴ)
= a * sqrt(sec^2ϴ)
=asecϴ

Question 15

sqrt (x^2 - a^2)

Answer 15

Trig Sub
x = asecϴ
= sqrt (a^2sec^2ϴ - a^2)
= sqrt (a^2(sec^2ϴ-1)
= a * sqrt (tan^2ϴ)
=atanϴ

Question 16

Midpoint Rule

Answer 16

x(i) = a + iΔx
∫ f(x) dx = Δx [f(x1)+f(x2)+…+f(xn)]
x(bar)(i) = x(i-1) + x(i) / 2
Δx = b-a / n

Question 17

Trapezoidal Rule

Answer 17

A(t) = Δx/2 [F(x(i-1)) + F(x(i))]
or
=Δx/2 [f(x0) + 2f(x1) + 2f(x2) + 2f(x3) +…+ f(xn)]

Question 18

∫ sqrt (1-cosx) dx

Answer 18

sin^2 (x/2) = 1 - cos2t /2 => sin^2 (x/2) = 1 - cosx /2
I = ∫ sqrt (2sin^2 (x/2)) dx
= sqrt 2 ∫ sin (x/2) dx
= sqrt 2 ∫ -2cos (x/2) +C
= -2sqrt(2) * cos(x/2) +C

Question 19

∫ sqrt ( 1-cosx) dx

Answer 19

Multiply by both top and bottom by Reciprocal
= sinx / sqrt (1+cosx) dx
u = 1 + cosx
du = -sinx dx => sinx dx = -du
I = ∫ -du / sqrt(u)
= ∫ u^-1/2 du
= -u^1/2 / 1/2 +C
= -2sqrt (1+cosx) +C

Question 20

∫ tan^3xsecx dx

Answer 20

U-Sub
Split apart first
tan^2x = sec^2x-1
Distribute sec^2x
∫ sec^2xtanxsecx dx - ∫ tanxsecx dx
u = secx
du = secxtanx dx
I = ∫ u^2 du - secx +C
= 1/3u^3 -secx +C
=1/3sec^3x-secx +C

Question 21

∫ x^-2 + x^-3 / x^-1 + 16x^-3 dx

Answer 21

U-Sub
Multiply Every Term by x^3 to remove negative exponents
∫ (1/x^2 + 1/x^3) / (1/x) + (16/x^3) dx
Multiply by reciprocal
∫ x + 1 / x^2 + 16 dx
Split apart
u = x^2 +16
du = 2x dx
xdx = du/2
I = 1/2 ∫ du/u + 1/4tan^-1(x/4) +C
= 1/2 ln(x^2 +16) + 1/4tan^-1(x/4)+C

Question 22

∫ dx / x(ln^2x+2lnx+2)

Answer 22

Multiple U-Sub
u = lnx
du = 1/x dx
Complete the square in the denominator
∫ du / (u^2 + 2u +2)
u^2+2u+2 = u^2+2u+1+1 = (u+1)^2 +1
∫ du / (u+1)^2 +1
v = u +1
dv = du
∫ dv / (v^2 +1)
Unsub , tan^-1v+C
Unsub, tan^-1 (u+1)
= tan^-1(1+lnx)
Evaluate the definite Integral fro [1/e,1]

Question 23

∫ 2x^2 + 3x +26 / (x-2)(x^2+16) dx

Answer 23

Partial Fraction Problem
= A / (x-2) + Bx+C / (x^2+16)
Multiply to remove common denominator
2x^2 + 3x +26 = A(x^2+16) + Bx+C(x-2)
Set x = to something that eliminates either A or B
For ex. x=2 ==> 40 = A(4+16)
A=2
X=0 ==> 26 = 2(16)+ c(-2)
C= 3
3 = -2B +C ==> 3 = -2B + 3
B = 0
I = ∫ ( 2 / (x-2) + 3 / (x^2+16)) dx
Split into 2 integrals
= 2ln(x-2) + 3/4tan^-1(x/4) +C

Question 24

∫ 3 / x^2+16 dx

Answer 24

U-Sub
3 * ∫ 1/x^2+16
u = x/4 , du = 1/4dx
= 3 ∫ 1 / 4(u^2 +1) du
= 3 * 1/4 ∫ 1 / u^2 +1
= 3/4tan^-1(u)
=3/4tan^-1(x/4) +C

Question 25

∫ 1 / x(x+3) dx

Answer 25

Partial Fractions
1 / x(x+3) = A / x + B / (x+3)
Multiply to remove denominator
1 = A(x+3) +Bx
if x = 0
1 = A(3) ==> A =1/3
if x = -3
1 = -3B ==> B = -1/3
I = ∫ ( 1/3 / x + (-1/3) / x+3 )dx
= 1/3 ∫ dx/x - 1/3 ∫dx/x+3
u = x+3
du = dx
= 1/3ln(x) - 1/3ln(x+3) +C

Question 26

x^4 + 2x^3 +x / (x^2-1)

Answer 26

Partial Fraction
Divide x^2 -1 into (x^4 +2x^3+x)
Which gives : x^2 +2x +1 + (3x+1/x^2-1)
3x+1 / (x-1)(x+1) = A / x-1 + B / x+1
Multiply to remove denominator
3x +1 = A(x+1) + B(x-1)
if x = -1 , -2 = B(-2)
B=1
if x=1 , 4 =A(2)
A=2
x^4+2x^3+x / x^2-1 = x^2 +2x+1 + (2 / x-1) + (1 / x+1)

Question 27

∫ 21x^3 / x^3-x^2-12x

Answer 27

Partial Fraction
Simplify top and bottom, factor denominator
= 21x / (x-4)(x+3) = A / (x-4) + B (x+3)
21x = A(x+3) + B(x-4)
x=-3 , -63 = B(-7)
B = 9
x=4 , 84 = A(7)
A =12
I = ∫ 12 / x-4 + 9 / x+3 dx
= 12 ∫ dx/x-4 + 9 ∫dx/x+3
=12ln(x-4) + 9ln(x+3) +C

Question 28

Improper Integrals

Answer 28

If the limit exists, we say that the integral converges, otherwise we say that it diverges.

Question 29

Comparison Theorem

Answer 29

Suppose that f and g are continuous functions w/
f(x) ≥ g(x) ≥ 0 , for x ≥ a
a) if ∫ f(x) dx is convergent for [a, infinity]
then ∫ g(x) dx is also convergent for [a, infinity]
Finite number = converges
Infinity = diverges

Question 30

∫ 1 / (x+e^x) dx ; a >0 from [a , ∞]

Answer 30

Comparison Theorem
1st Try:
1/x+e^x < 1/x => ∫ 1/x dx from [a , ∞] diverges
2nd Try:
1/x+e^x < 1/e^x => ∫1/e^x dx
= lim b-> ∞ ∫ e^-x dx [a,b]
= lim b-> ∞ -e^-x ] from a,b
= lim b-> ∞ (-e^-b - (-e^-a)
b = infinity
= e^-a , a is a finite number
therefore the function converges.