Integration is tricky because integrands come in so many shapes, so you choose a method that fits the shape.
Substitution replaces an inner function with u and turns the integral into a simpler one, while integration by parts runs the product rule backward to trade a hard integral for an easier one.
A square root suggests a trigonometric substitution and a fraction suggests partial fractions: you read the method off the shape.
In the first picture, move the upper limit t and the area on the x side and the area on the u side always come out to the same value, sin(t²); in the second, the area under xeˣ matches the value found by parts.
Integration by Substitution
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Left: area under 2x cos(x²) from 0 to t. Right: area under cos u from 0 to t². The two are equal
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🔄 From an x-integral to a u-integral
①In 2x cos(x²), x² sits inside the cosine and its derivative 2x is multiplied in front
②With u = x² we get du = 2x dx, so the whole of 2x cos(x²) dx becomes cos u du
③The interval changes too: as x goes from 0 to t, u goes from 0 to t². That is why the two areas in the picture always equal sin(t²)
④So the antiderivative is sin u + C = sin(x²) + C
Substitution
∫ f(g(x)) · g′(x) dx = ∫ f(u) du (u = g(x))
Setting the inner function to u = g(x) gives g′(x) dx = du. For a definite integral, change the limits to values of u as well
Integration by Parts
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Area under y = xeˣ from 0 to t, computed by parts
1.5
Integration by parts
∫ u dv = uv − ∫ v du
Integrate both sides of the product rule (uv)′ = u′v + uv′ and rearrange. Choose as u the factor that gets simpler when differentiated, and as dv the factor that is easy to integrate
🧩 Choosing u
①A common order of preference for u is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential
②In ∫xeˣ dx, take the polynomial x as u and eˣ dx as dv, so du = dx and v = eˣ
③Then ∫xeˣ dx = xeˣ − ∫eˣ dx = xeˣ − eˣ + C, and from 0 to t this gives (t − 1)eᵗ + 1, the value shown under the picture
Methods by Shape
Trigonometric substitution
√(a² − x²) → x = a sin θ, √(x² − a²) → x = a sec θ, √(x² + a²) → x = a tan θ
Choose the substitution from the expression under the root. Identities such as 1 − sin²θ = cos²θ are what make the root disappear
Partial fractions
1(x-a)(x-b) = Ax-a + Bx-b
Split a rational function whose denominator is a product of linear factors into partial fractions and integrate each one as a logarithm. Multiply both sides by (x − a)(x − b) and set x = a and x = b to find A and B
💡 How to choose
①An inner function together with its derivative points to substitution
②A product of two different kinds of function points to integration by parts
③A root containing something like a² − x² points to trigonometric substitution
④A fraction whose denominator factors points to partial fractions
⑤Differentiate your answer to check that the integrand comes back
Integrals to Know
Exponential and 1/x
∫ eˣ dx = eˣ + C, ∫ 1x dx = ln|x| + C
The integral of eˣ is eˣ itself, and the integral of 1/x is ln|x|; the absolute value covers the interval where x < 0 as well
Trigonometric integrals
∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C
Because (cos x)′ = −sin x, the integral of sin x carries a minus sign
Integrals of sec² and csc²
∫ sec²x dx = tan x + C, ∫ csc²x dx = −cot x + C
These are (tan x)′ = sec²x and (cot x)′ = −csc²x read backward
Worked Examples
Example 1
Find the indefinite integral ∫ 2x cos(x2) dx.
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Let the inner function x2 be u, so du = 2x dx.
u = x2, du = 2x dx
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Integrate with respect to u, then switch back to x.
∫ cos u du = sin u + C = sin(x2) + C
▸ sin(x2) + C
When the derivative of an inner function appears as a factor, substitution will work.
Example 2
Find the indefinite integral ∫ x ex dx.
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Let u = x and dv = ex dx, so du = dx and v = ex.
u = x, dv = ex dx ⇒ du = dx, v = ex
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Substitute into ∫ u dv = uv − ∫ v du.
x ex − ∫ ex dx = x ex − ex + C
▸ (x − 1)ex + C
Taking the polynomial as u lowers its degree each time it is differentiated, so the integral gets easier. Taking eˣ as u instead would raise the power of x and make things worse.