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B Veitch Calculus 2 Derivative and Integral Rules

1. Derivative Formulas

(a) Common Derivatives

i.d

dx(c) = 0

ii.d

dx(f ± g) = f ′ ± g′

iii.d

dx(x) = 1

iv.d

dx(kx) = k

v. Power Rule:d

dx(xn) = nxn−1

vi.d

dx([f(x)]n) = n[f(x)]n−1 · f ′(x)

vii. Product Rule: (f g)′ = f ′g + fg′

viii. Quotient Rule:

(f

g

)′=f ′g − fg′

g2

ix. Chain Rule: (f(g(x)))′

= f ′(g(x)) · g′(x)

x.d

dx(ax) = ax ln a

xi.d

dx(ex) = ex

xii.d

dx(lnx) =

1

x

xiii.d

dx(loga x) =

1

x ln a

xiv.d

dx

(ef(x)

)= f ′(x) · ef(x)

xv.d

dx(ln(f(x))) =

1

f(x)· f ′(x)

xvi.d

dx(sinx) = cosx

xvii.d

dx(cosx) = − sinx

xviii.d

dx(tanx) = sec2 x

xix.d

dx(secx) = secx tanx

xx.d

dx(cscx) = − cscx cotx

xxi.d

dx(cotx) = − csc2 x

xxii.d

dx

(sin−1 x

)=

1√1− x2

xxiii.d

dx

(cos−1 x

)=

−1√1− x2

xxiv.d

dx

(tan−1 x

)=

1

1 + x2

xxv.d

dx

(cot−1 x

)=−1

1 + x2

xxvi.d

dx

(sec−1 x

)=

1

x√x2 − 1

xxvii.d

dx

(csc−1 x

)=

−1

x√x2 − 1

2. L’Hospitals Rule

(a) If limx→a

f(x) = 0 and limx→a

g(x) = 0, then

limx→a

f(x)

g(x)= lim

x→a

f ′(x)

g′(x), g′(x) 6= 0

(b) If limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞, then

limx→a

f(x)

g(x)= lim

x→a

f ′(x)

g′(x)

(c) Indeterminate Form of Type 0 · ∞

Given limx→a

f(x)g(x), you do the following to put it in the form0

0or∞∞

i. Rewrite f(x)g(x) asf(x)

1/g(x)

ii. Rewrite f(x)g(x) asg(x)

1/f(x)

(d) Indeterminate Form of 00, ∞0, or 1∞

Given limx→a

f(x)g(x), rewrite the limit as

limx→a

eg(x) ln(f(x))

1

B Veitch Calculus 2 Derivative and Integral Rules

then take the limit of the exponent

limx→a

g(x) ln(f(x))

This should put the limit in the Indeterminate Form of Type 0 · ∞

3. Integrals

(a) Common Integrals

i.

∫k dx = kx+ C

ii.

∫xn dx =

xn+1

n+ 1+ C, n 6= −1

iii.

∫1

xdx = ln |x|+ C

iv.

∫1

kx+ bdx =

1

kln |kx+ b|+ C

v.

∫ex dx = ex + C

vi.

∫ekx dx =

1

kekx + C

vii.

∫ekx+b dx =

1

kekx+b + C

viii.

∫ax dx =

ax

ln a+ C

ix.

∫akx dx =

1

k ln aakx + C

x.

∫akx+b dx =

1

k ln aakx+b + C

xi.

∫lnx dx = x lnx− x+ C

xii.

∫cosx dx = sinx+ C

xiii.

∫sinx dx = − cosx+ C

xiv.

∫sec2 x dx = tanx+ C

xv.

∫csc2 x dx = − cotx+ C

xvi.

∫secx tanx dx = secx+ C

xvii.

∫cscx cotx dx = − cscx+ C

xviii.

∫tanx dx = ln | secx|+ C

xix.

∫cotx dx = ln | sinx|+ C

xx.

∫cscx dx = ln | cscx− cotx|+ C

xxi.

∫secx dx = ln | secx+ tanx|+ C

xxii.

∫1√

a2 − x2dx = sin−1

(xa

)+ C

xxiii.

∫1

a2 + x2dx =

1

atan−1

(xa

)+ C

xxiv.

∫− 1

a2 + x2dx =

1

acot−1

(xa

)+ C

xxv.

∫1

x√x2 − 1

dx = sec−1 (x) + C

xxvi.

∫− 1

x√x2 − 1

dx = csc−1 (x) + C

4. Integration Techniques

(a) u Substitution

Given

∫ b

a

f(g(x))g′(x) dx,

i. Let u = g(x)

ii. Then du = g′(x) dx

iii. If there are bounds, you must change them using u = g(b) and u = g(a)∫ b

a

f(g(x))g′(x) dx =

∫ g(b)

g(a)

f(u) du

(b) Integration By Parts ∫u dv = uv −

∫v du

Example:

∫x2e−x dx

2

B Veitch Calculus 2 Derivative and Integral Rules

u = x2 dv = e−x dx

du = 2x dx v = −e−x∫x2e−x dx = −x2e−x −

∫−2xe−x dx

You may have to do integration by parts more than once. When trying to figure out what to

choose for u, you can follow this guide: LIATE

L Logs

I Inverse Trig Functions

A Algebraic (radicals, rational functions, polynomials)

T Trig Functions (sinx, cosx)

E Exponential Functions

(c) Products of Trig Functions

i.

∫sinn x cosm x dx

A. m is odd (power of cosx is odd). Fac-

tor out one cosx and place it in front

of dx. Rewrite all remaining cosx as

sinx by using cos2 x = 1−sin2 x. Then

let u = sinx and du = cosx dx

B. n is odd (power of sinx is odd). Fac-

tor out one sinx and place it in front

of dx. Rewrite all remaining sinx as

cosx by using sin2 x = 1−cos2 x. Then

let u = cosx and du = − sinx dx

C. If n and m are both odd, you can

choose either of the previous methods.

D. If n and m are even, use the following

trig identities

sin2 x =1

2(1− cos(2x))

cos2 x =1

2(1 + cos(2x))

sin(2x) = 2 cosx sinx

ii.

∫tann x secm x dx

A. m is even (power of secx is even). Fac-

tor out one sec2 x and place it in front

of dx. Rewrite all remaining secx as

tanx by using sec2 x = 1 + tan2 x.

Then let u = tanx and du = sec2 x dx

B. n is odd (power of tanx is odd). Fac-

tor out one secx tanx and place it

in front of dx. Rewrite all remain-

ing tanx as secx by using tan2 x =

sec2 x − 1. Then let u = secx and

du = secx tanx dx

C. If n odd and m is even, you can use

either of the previous methods.

D. If n is even and m is odd, the previous

methods will not work. You can try to

simplify or rewrite the integrals. You

may try other methods.

(d) Partial Fractions

Use this method when you are integrating

∫p(x)

q(x)dx, the degree of p(x) must be smaller than

the degree of q(x). Factor the denominator q(x) into a product of linear and quadratic factors.

There are four scenarios.

3

B Veitch Calculus 2 Derivative and Integral Rules

Unique Linear Factors: If your denominator has unique linear factors

x

(2x− 1)(x− 3)=

A

2x− 1+

B

x− 3

To solve for A and B, multiply through by the common denominator to get

x = A(x− 3) +B(2x− 1)

You can find A by plugging in x =1

2. Find B by plugging in x = 3.

Repeated Linear Factors: Every power of the linear factor gets its own fraction (up to the

highest power).

x

(x− 3)(3x+ 4)3=

A

x− 3+

B

3x+ 4+

C

(3x+ 4)2+

D

(3x+ 4)3

Unique Quadratic Factor:

3x− 1

(x+ 4)(x2 + 9)=

A

x+ 4+Bx+ C

x2 + 9

To solve or A, B, and C, multiply through by the common denominator to get

3x− 1 = A(x2 + 9) + (Bx+ C)(x+ 4)

Repeated Quadratic Factor: Every power of the quadratic factor gets its own fraction (up

to the highest power).

2x

(3x+ 4)(x2 + 9)3=

A

3x+ 4+Bx+ C

x2 + 9+

Dx+ E

(x2 + 9)2+

Fx+G

(x2 + 9)3

(e) Trig Substitution

If you see Substitute Uses the following Identity√a2 − x2 x = a sin(θ) 1− sin2(θ) = cos2(θ)√a2 + x2 x = a tan(θ) 1 + tan2(θ) = sec2(θ)√x2 − a2 x = a sec(θ) sec2(θ)− 1 = tan2(θ)

Example:

∫x3√

16− x2dx Let x = 4 sin θ, dx = 4 cos θ dθ. So x3 = 64 sin3 θ

∫x3√

16− x2dx =

∫64 sin3 θ√

16− 16 sin2 θ· 4 cos θ dθ =

∫64 sin3 θ · 4 cos θdθ

4 cos θ=

∫sin3 θ dθ

Now you have an integral containing powers of trig functions. You can refer to that method to

solve the rest of this integral. ∫sin3 dθ =

cos3 θ

3− cos θ + C

To get back to x, we need to use a right triangle with the original substitution x = 4 sinx.

4

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