
StudySquare
Methods of integration for Edexcel A-level Maths







This page covers the following topics:
1. Integration by parts
2. Integration by substitution
3. Integrating partial fractions
Integration by parts is a special method of integration which is used when two functions are being multiplied. They are defined as u and v accordingly, and the given formula is used to evaluate the interval.

Integration by substitution is a method to evaluate integrals by changing the variables. This method can be used when the integral is of the given form.

When the integrand is in the form of a proper fraction, the integral should be rewritten as partial fractions and then integrated using the given result.

1
Evaluate ∫1/(x + 5)dx.
∫1/(x + 5)dx = ln|x + 5| + c, where c is the constant of integration.
2
Use integration by parts to find ∫(5 + x)cos(x)dx.
Let u = 5 + x and dv = cos(x)dx.
Then, du = dx and v = sin(x).
∫ (5 + x)cos(x)dx = (5 + x)sin(x) − ∫sin(x)dx =
= (5 + x)sin(x) + cos(x) + c, where c is the constant of integration
(5 + x)sin(x) + cos(x) + c
3
Determine the following integral: ∫3x²sin(x³ − 10x) − 10sin(x³ − 10x)dx.
The integral can be rewritten as ∫(3x² − 10)sin(x³ − 10x)dx. To integrate by substitution, let u = x³ − 10x and du = (3x² − 10)dx. So, ∫(3x² − 10)sin(x³ − 10x)dx = ∫sin(u)du = −cos(u) + c = −cos(x³ − 10x) + c, where c is the constant of integration.
4
Evaluate ∫xeˣ.
Let u = x and dv = eˣdx. Then, du = 1dx and v = eˣ. So, ∫xeˣ = xeˣ − ∫eˣdx = xeˣ − eˣ + c, where c is the constant of integration.
5
Integrate the area of the given rectangle.
The integral is ∫15x²sin(8 − 5x³)dx. To integrate by substitution, let u = 8 − 5x³ and du = −15x²du. So, ∫15x²sin(8 − 5x³)dx = ∫−sin(u)dx = cos(u) + c = cos(8 − 5x³) + c, where c is the constant of integration.

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