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Wash U - Study Guide - Midterm

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Calc II – Study Guide – Exam 2

Integration by Parts

∫ ��(��)��′(��)���� = ��(��)��(��) − ∫ ��′(��)��(��)����

• For a function composed of products, integration by parts can be a useful tool • Steps:

o Start by defining f(x) and g’(x)

o Find f’(x) by taking derivative of f(x) and find g(x) by taking the antiderivative of g’(x)

o Plug into equation and solve for the integral of f’(x)g(x)

• Can also be in the form of:

∫ ������ = ���� − ∫ ������

• Example: ∫ �� cos �� ����

o f(x) = x

o f’(x) = 1

o g’(x) = cos x

o g(x) = sin x

o ∫ �� cos �� ���� = �� sin �� − ∫ 1 sin �� ���� = �� sin �� + cos �� + ��

• When choosing f(x) and g’(x):

o Look for the function with a degree that would be reduced by differentiation, this will be f(x)

▪ Ex: x, x2, etc.

▪ Functions we do not know the antiderivative of (i.e. ln(x))

o Look for a function with a degree that won’t change due to differentiation, this will be g’(x)

▪ Ex: cos(x), sin(x), ex, etc.

“Trig Tricks”

• For integrals involving two trig functions, there are some “trig tricks” useful for antidifferentiation

• Involves using u-sub for the odd-powered function:

o If the power of sin(x) is odd, use u = cos(x)

o If the power of cos(x) is odd, use u = sin(x)

o If both powers are even, use double-angle identities

• Useful trigonometric identities:

For…

Use…

Translating sin(x) to cos(x)

sin2 �� = 1 − cos2 ��

Translating cos(x) to sin(x)

cos2 �� = 1 − sin2 ��

Reducing the power of sin(x)

sin2 �� =12(1 − cos 2��)We also discuss several other topics like cumbium

Reducing the power of cos(x)

cos2 �� =12(1 + cos 2��)

• Ex 1: ∫ sin2 �� ����

o sin2 �� =12(1 − cos 2��)

▪ First, reduce power of sin

o ∫12(1 − cos 2��) ����

▪ Substitute back into integral

o �� = 2�� and ���� = 2����

▪ Use u-sub to simplify cos(2x)

o12∫(1 − cos ��)����2

▪ Put integral in terms of u

o14(�� − sin ��) + ��

▪ Integrate

o14(2�� − sin 2��) + ��

▪ Put back in terms of x

• Ex 2: ∫ cos2 �� ����

o cos2 �� =12(1 + cos 2��)

o ∫12(1 + cos 2��) ����

o �� = 2�� and ���� = 2����

o12∫(1 + cos ��)����2

o14(�� + sin ��) + ��

o14(2�� + sin 2��) + ��

• Ex 3: ∫ sin3 �� cos3 �� ����

o u = cos(x) and du = -sin(x) dx

▪ Either sin(x) or cos(x) could be chosen for the u-sub

o ∫ sin3 �� ��3(����

− sin��)

▪ Put in terms of u and du

o ∫ − sin2 �� ��3����

▪ Simplify

o − ∫(1 − cos2 ��)��3����

▪ Substitute for sin(x) so that it can be written in terms of u

o − ∫(1 − ��2)��3����

▪ Put equation fully in terms of u

o − ∫ ��3 − ��5����

▪ Distribute

o −14��4 +16��6 + �� If you want to learn more check out comm 371 exam 2

▪ Take the antiderivative of the equation

o −14cos4 �� +16cos6 �� + �� If you want to learn more check out mqau

▪ Put back in terms of x

Trig Sub

• For equations with quadratic functions in inconvenient places, this technique can be helpful

• A useful table:

For:

x Substitution

dx Substitution

Trig Formula

a2 – x2

x = a sinθ

dx= a cosθ dθ

1 - sin2θ = cos2θ

a2 + x2

x = a tanθ

dx = a sec2θ dθ

1 + tan2θ = sec2θ

x2 – a2

x = a secθ

dx = a secθtanθ dθ

sec2θ - 1 = tan2θ

• Ex: ∫√��2−4

��3 ����

o The quadratic in question can be written as x2- 22, which calls for a secθ substitution, where a is 2

o Let x = 2 secθ and dx = 2secθtanθ dθ

o ∫√(2 sec θ)2−22

(2 sec��)32 sec �� tan �� ����

▪ Substitute x = 2 secθ and dx = 2 secθtanθ dθ

o ∫√4 sec2 ��−4

8 sec3 ��2 sec �� tan �� ����

▪ Expand the exponents

o ∫√4(sec2 ��−1)

8 sec3 ��2 sec �� tan �� ����

▪ Factor out a 4 so a substitutable equation is evident o ∫√4(tan2 ��)

8 sec3 ��2 sec �� tan �� ����

▪ Substitute sec2θ - 1 = tan2θ

o ∫2 tan ��

8 sec3 ��2 sec �� tan �� ����

▪ Take the square root

o12∫tan2 ��

sec2 ������

▪ Simplify the terms and pull out the ½ constant o12∫sin2 ��

cos2 �� 1

cos2 ��

����

▪ Break down tan2θ into sin2θ/cos2θ and sec2θ into 1/cos2θ

o12∫ sin2 �� ����

▪ Simplify equation (this integral was completed earlier)

o14�� −18sin 2�� + ��

▪ Solve (same as 14(�� + sin ��) + ��)

o �� = arcsec��2

▪ Solve for theta

o14(arcsec��2) −18sin 2 (arcsec��2) + �� Don't forget about the age old question of statuatory law

• Sometimes, completing the square is necessary to get to a substitutable point o ����2 + ���� + �� → ��(�� + ��)2 + ��

o �� =��2��

o �� = �� −��2

4��

o Ex: 4��2 + ��

▪ 4 (��2 +14��) = 4 (��2 +14�� +164−164)

▪ 4 ((�� +18)2−164)

▪ �� =18tan ��

• When solving for definite integrals, convert back to x before plugging in boundaries Partial Fractions

• For integrating rational functions, this technique can be used

• It involves setting one function equal to two added together

• Ex: ∫1

��2−1����

o Factor denominator (�� − 1)(�� + 1)

o Find A and B so that 1

��2−1=��

��−1+��

��+1

o Do this by multiplying A and B by factors that would make the denominators equal, set equal to original numerator: ��(�� + 1) + ��(�� − 1) = 1

o Solve for A and B using a system of equations (A = 1/2 and B = -1/2) o Substitute back into integral and solve:

��−1−(12)

1

▪ ∫ (

2

��+1) ����

▪12ln |�� − 1| −12ln|�� + 1| + ��

• Can result in three or more factors (A, B, C, etc.)

o Ex: ��3 − 4�� = ��(�� + 2)(�� − 2)

• If factors are repeated, add an extra term

o Ex: 1

��(��−2)2 =����+��

��−2+��

(��−2)2

• If factors are non-linear, add an extra term

o Ex: 1

��2(��−1)(��+4)=����+����2 +��

��−1+��

(��+4)

• If the numerator is a higher degree than the denominator, use long division to simplify o Ex: ��3+2

��2−1= �� +��+2

��2−1

Integration Strategies

Technique Don't forget about the age old question of Why factories built in cities as close to markets and transportation?

Useful for

Ex

U-Sub

Nested functions

∫ cos 2�� ����

Integration by Parts

Products

∫ ������ ����

“Trig Tricks”

Powers of sin(x)/cos(x)

∫ cos3 �� sin2 �� ����

Trig Sub

Quadratic functions in inconvenient places Don't forget about the age old question of ut bas

∫1

√1 − ��2����

Partial Fractions

Rational functions

∫�� + 2

��2 − 3�� + 2����

References: