Integration

Geometry

Sometimes you can integrate a function easily by taking it's geometry.

For example, int1-3(y=x-1)dx is just a triangle so the answer is (2*2) / 2

Also int0-1(y=sqrt(1-x^2))dx is the top half of a circle x^2 + y^2 = 1. Given the range you are taking 1/4 of the area of the circle so (pi(1)^2)/4 = pi/4.

The Fundamental Theorem of Calculus

If f is continuous on the interval [a, b] and f(x) = F'(x), then inta-b(f(x)dx) = F(b) - F(a)

The function F(x) is known as the anti derivative of f(x)

Indefinite Integrals

When you find anti-derivatives you just use the existing differentiation rules but in reverse.

example: f(x) = 2x has a family of anti derivatives F(x) = x^2 + C

Constant

∫kdx = kx + C
∫dx = x + C

Power functions

∫x^ndx = [x^(n+1)/(n+1)] + C when n != -1

Keep in mind all the rules of fractions and exponents so you can jump around

Example:

1/x^(9/4)
x^(-9/4)
(1/(-9/4 + 4/4))x^(-9/4 + 4/4)
(1/(-5/4))x^(-5/4)
(-4/5)x^(-5/4)

-4/(5x^(5/4))

Sums, differences, constant multiples

∫(f(x) + g(x))dx = ∫f(x)dx + ∫g(x)dx
∫(f(x) - g(x))dx = ∫f(x)dx - ∫g(x)dx
∫cf(x)dx = c∫f(x)dx

Exponential and logarithmic

∫e^x dx = ex + C
∫a^x dx = (a^x)/ln(a) + C
∫1/x dx = ln(|x|) + C

Trigonometric

∫sinx dx = -cosx + C
∫cosx dx = sinx + C
∫sec^2(x) dx = tanx + C
∫secxtanx dx = secx + C
∫-cotxcscx dx = cscx + C
∫-csc^2x dx = cotx + C
∫1/(sqrt(1-x^2)) dx = sin^-1(x) + C = arcsinx + C
∫1/(1+x^2) dx = tan^-1(x) + C = arctanx + C

More involved:
∫cscx dx = -ln(|cscx+cotx|) + C
∫secx dx = ln(|secx+tanx|) + C

Some of them have a trick where you multiply by "1" and do u sub… it's rough 😞

∫cscx dx

∫(csc^2x+cscxcotx)/(cscx + cotx) dx

u = cscx+cotx
du = -cscxcotx - csc^2x dx

∫((csc^2x+cscxcotx)/u)(du/(-1(cscxcotx+csc^2x)))

-1 * ∫1/u du

-ln(|u|) + c

-ln(|cscx+cotx|)+C

Initial Value Problems

Sometimes you are provided with an initial value(s) so that the family of anti derivatives can be narrowed down to a single anti derivative.

Find f(x) if f'(x) = cos(x) + sin(x) and f(0) = 5.

Initial values are also helpful because you can graph the resulting function.

Average value of a function

aka average slope of all tangent lines of the derivative

signed area/width -> (∫[interval]f(x)dx)/interval

avg value of f(x) on interval [a, b]
take (∫[a to b]f(x)dx)/b - a
(F(b) - F(a))/b - a

U Substitution

Some functions can not be integrated using the above rules but you can use a process called u-substitution.

This can only work if the function has one "chunk" that is the derivative of another "chunk" of the function. For example: in ∫3x^2(x^3 + 5)^7 dx 3x^2 is the derivative of x^3 + 5. The chunk will almost always be the longest x expression that is composed inside another function, like x^7 in this case.

1. Define a new variable u to be the chunk of the function that has it's derivative in the function, so in this case u = x^3 + 5
2. Take the derivative of u. du = 3x^2 dx
3. If du does not exist exactly in the remaining function, you have to manipulate du and/or u algebraically until you have exactly what you need for the substitution to work out. See other examples for this.
4. Now that you have u and du, substitute into the original function: ∫(u^7)du
5. Integrate this new function: ((u^8)/8) + C The du is removed here because you inverted the chain rule behind the scenes.
6. Back substitute u and simplify if you can: (((x^3 + 5)^8)/8) + C

Check the definite integral example to see special steps for definite integrals

Examples

Rearrange the du expression

∫(x^3)/sqrt(1 - x^4) dx

u = 1 - x^4
du = -4x^3 dx

-4x^3 does not exist in the original expression

(-1/4)du = x^3 dx

now it does :), so substitute

∫((-1/4)du)/sqrt(u)

-1/4∫(1/u^1/2)du

-1/4∫(u^(-1/2))du

now integrate

-1/4(u^1/2 / 1/2) + C

simplify

-1/4(2)u^1/2 + C

back substitute u

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

Rearrange the u expression

∫xsqrt(x + 2) dx

u = x + 2
du = dx

rearrange u so you can substitute

x = u - 2

now substitute

∫(u - 2)sqrt(u) du

simplify

∫(u - 2)u^1/2

∫u^3/2 - 2u^1/2 du

integrate

(u^5/2 / 5/2) - (2c^3/2 / 3/2) + C

simplify

(2/5)u^5/2 - (4/3)u^3/2 + C

back substitute

(2/5)(x + 2)^5/2 - (4/3)(x + 2)^3/2 + C

Trig

∫sinx/cos^3x dx

u = cosx
du = -sinx dx -> -du = sinx dx

∫-du/u^3 = -∫1/u^3 du = -∫u^-3 du

integrate

-(u^-2 / -2) + C = (1/2)u^-2 + C = (1/2u^2) + C

back substitute

(1/(2cos^2x)) + C

Definite integral

As soon as you get rid of the x variables and replace them with u variables, you need to adjust the limits by plugging them in for x in the u expression.

∫[0 to 2]2x(x^2 + 4)^2 dx

u = x^2 + 4
du = 2xdx -> du/2x = dx

∫[0 to 2]u^2 du // dx canceled out

adjust limits

u = 0^2 + 4 = 4 -> lower limit is 4
u = 2^2 + 4 = 8 -> upper limit is 8

∫[4 to 8]u^2 du

evaluate

(8^3 / 3) - (4^3 / 3)

448/3

Integration by parts

Really just the reverse of the product rule for derivatives

This expression is true:

∫u dv = uv - ∫v du

or written as

∫f(x)g'(x) = f(x)g(x) - ∫g(x)f'(x)

And you can use it to rewrite integrals so they are solvable

∫xe^x dx

u  = x
dv = e^x dx
du = 1 dx <- derivative of u
v  = e^x <- integrate dv (+C does not matter here)

∫xe^x dx = xe^x - ∫e^x dx

now solve the new equation

xe^x - e^x + C

If you are struggling to pick u and dv, follow LIATE in order and take something of that type for u. u should get simpler when you take the derivative and dv should not get more complicated. dv always gets the dx.

• L logs
• I inverse trig
• A algebraic
• T trig
• E exponential

Solving integration by parts problems

• Sometimes you have to do integration by parts multiple times until you get something you can integrate.

• Check to see if it makes sense to do a usub first. It might make the integration by parts easier

• If you only have one function, sometimes you can just introduce a 1 to make the problem solvable. ∫ln(x)dx -> ∫1*ln(x)dx. 1 is the derivative of x.

• If you do integration by parts multiple times and end up with the original integral in your formula, you can move it over: original = formula - original so 2original = formula so original = formula/2. This is called boomerang integral.

• For definite integrals, the formula gets that line symbol with lower and upper limit on it to show you would calculate both and subtract lower from upper. this - that |. You can't really type it lol so look at an example. Also don't add +c obviously.

Method of partial fractions

The integral of some rational functions can be found by splitting the integrand into partial fractions. You want to get the multiplication out of the denominator. The hint for this is that substitution is not going to work:

∫ x - 5 / [(2x - 3)(x - 1)] dx

Partial fractions is a technique from algebra where you split one fraction into two and then add them together with a common denominator to find a system of equations that will give you the constant multiples that are the "cost" of splitting the fractions:

A/2x-3 + B/x-1

A(x-1)/[(2x-3)(x-1)] + B(2x-3)/[(2x-3)(x-1)]

distribute and add

(A + 2B)x - (A + 3B)/[(2x-3)(x-1)]

the original numerator is x - 5 and the new one is (A + 2B)x - (A + 3B)

so now you know:
A + 2B = 1
A + 3B = 5

Solve the system of equations
b = 4
a = -7

Now you know A and B, so you can rewrite the original integral and break it into two solvable integrals:

∫ -7/(2x-3)dx + ∫ 4/(x-1)dx

-7/2∫ 2/(2x-3)dx + 4∫ 1/(x-1)dx

(-7/2)ln(|2x-3|) + 4ln(|x-1|) + C

Keep in mind you may need to factor the denominator first to produce a denominator where you can split the fractions. x^2 + x = x(x + 1) = A/x + B/x+1

Exceptional cases

If you have two factors and one or more is to a power, you have to add all fractions up to and including that power.

1/[(x+1)^2 * (x+3)^3]

A/x+1 + B/(x+1)^2 + C/x+3 + D/(x+3)^2 + E/(x+3)^3

If you are given an irreducible factor, you have to account for it in the numerator of the fraction over the irreducible factor.

x+1/[x(x^2+1)]

A/x + Bx+C/(x^2 + 1)

Trig Integrals

Trig integrals often have an additional step you can preform to clean up the integrand and create a more straight forward integration problem.

Typically you will see a trig value raised to an odd power, and you can use a pythagorean identity to remove all but one of them and produce a straight forward u-sub problem.

∫sin^4(x)cos^5(x)dx

∫sin^4(x)cos^4(x)cos(x)dx

we know cos^2 = 1 - sin^2 so therefore cos^4 = (1-sin^2)^2

∫sin^4(x)((1-sin^2(x))^2)cos(x)dx

u = sin(x), du = cos(x)dx

∫u^4((1-u^2)^2)du

FOIL it out

∫u^4(1 + u^4 - 2u^2)du

∫(u^4 + u^8 - 2u^6)du

1/5sin^5(x) + 1/9sin^9(x) - 2/7sin^7(x) + C

You can do the same thing with tan and sec using tan^2(x) + 1 = sec^2(x)

If none of the trig statements are raised to an odd power, you can use double angle identities. Found here. It's common you need to use it more than once when solving the problem.

Trig substitution

Sometimes an integral seems hard to solve and you need to do a clever type of u-sub where you plug in a trig function.

∫1/(1+x^2) dx

x = tan(Θ) or arctanx = Θ
dx = sec^2(Θ) dΘ

∫sec^2(Θ)/(1+tan^2(Θ)) dΘ

∫sec^2(Θ)/sec^2(Θ) dΘ

∫1 dΘ

Θ

tan^-1(x) + C

Pick the trig identity that fits your problem. 1 - x probs means sin, 1 + x probs means tan, etc…

∫sqrt(a^2 - x^2) dx -> x = asin(Θ)

∫sqrt(a^2 + x^2) dx -> x = atan(Θ)

∫sqrt(x^2 - a^2) dx -> x = asec(Θ)

If you pick the correct trig function but it's not working, you may need to multiply the trig function by a constant:

∫1/sqrt(4-x^2)

x = 2sin(Θ)
dx = 2cos(Θ)dΘ

∫2cosΘ/(sqrt(4)sqrt(1-sin^2Θ)) dΘ

...

arcsin(x/2) + C