Cambridge A-Level Maths 9709 - Pure Maths 3 (P3)

Integration: Techniques

This section covers various techniques for integrating functions. Mastering these techniques is crucial for solving a wide range of integration problems.

Basic Integration Rules

• $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$)
• $\int \frac{1}{x} dx = \ln|x| + C$
• $\int e^x dx = e^x + C$
• $\int \sin(x) dx = -\cos(x) + C$
• $\int \cos(x) dx = \sin(x) + C$

Integration by Substitution

This technique is used when the integrand contains a composite function.

1. Choose a suitable substitution for the inner function.
2. Calculate the derivative of the chosen substitution.
3. Substitute the substitution and its derivative into the integral.
4. Solve the resulting integral.
5. Substitute back to the original variable.

Example: $\int 2x \cos(x^2) dx$

Let $u = x^2$, then $du = 2x dx$.

$\int \cos(u) du = \sin(u) + C = \sin(x^2) + C$

Integration by Parts

This technique is used when the integrand is a product of two functions.

Formula: $\int u dv = uv - \int v du$

Choose $u$ and $dv$ such that $\int v du$ is simpler than $\int u dv$.

Example: $\int x e^x dx$

Let $u = x$ and $dv = e^x dx$. Then $du = dx$ and $v = e^x$.

$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$

Integration of Trigonometric Functions

Techniques involve using trigonometric identities to simplify the integrand.

Example: $\int \sin^2(x) dx$

Using the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

$\int \frac{1 - \cos(2x)}{2} dx = \frac{1}{2} \int (1 - \cos(2x)) dx = \frac{1}{2} (x - \frac{1}{2} \sin(2x)) + C = \frac{1}{2} x - \frac{1}{4} \sin(2x) + C$

Integration of Exponential and Logarithmic Functions

These often require specific techniques or the use of standard integration formulas.

Example: $\int e^{2x} dx$

Let $u = 2x$, then $du = 2 dx$, so $dx = \frac{1}{2} du$.

$\int e^u \frac{1}{2} du = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C = \frac{1}{2} e^{2x} + C$

Volumes of Revolution

This section deals with finding the volume of solids formed by rotating a two-dimensional region around a line.

Disk/Washer Method

Disk Method: When the region is rotated around a horizontal or vertical axis.

Volume = $\pi \int_a^b [f(x)]^2 dx$ (for rotation around the x-axis) or $\pi \int_a^b [f(y)]^2 dy$ (for rotation around the y-axis)

Washer Method: When the region is rotated around a line that is not parallel to the axis of rotation.

Volume = $\pi \int_a^b [R(x)^2 - r(x)^2] dx$ (where $R(x)$ is the outer radius and $r(x)$ is the inner radius)

Example: Find the volume of the solid formed by rotating the region bounded by $y = x^2$, $y = 0$, and $x = 2$ around the x-axis.

Volume = $\pi \int_0^2 (x^2)^2 dx = \pi \int_0^2 x^4 dx = \pi [\frac{x^5}{5}]_0^2 = \pi (\frac{32}{5}) = \frac{32\pi}{5}$

Shell Method

When the region is rotated around a vertical axis.

Volume = $2\pi \int_a^b x f(x) dx$ (where $f(x)$ is the height of the shell and $x$ is the radius of the shell)

Example: Find the volume of the solid formed by rotating the region bounded by $y = x^2$, $y = 0$, and $x = 2$ around the y-axis.

Volume = $2\pi \int_0^2 x (x^2) dx = 2\pi \int_0^2 x^3 dx = 2\pi [\frac{x^4}{4}]_0^2 = 2\pi (\frac{16}{4}) = 8\pi$

Differential Equations

This section introduces the concept of differential equations and methods for solving them.

Separable Equations

Equations that can be written in the form $f(y) dy = g(x) dx$.

Solve by integrating both sides.

Example: $\frac{dy}{dx} = \frac{x}{y}$

$y dy = x dx$

$\int y dy = \int x dx$

$\frac{y^2}{2} = \frac{x^2}{2} + C$

$y^2 = x^2 + 2C$

Linear Equations

Equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$

Solve using an integrating factor $\mu(x) = e^{\int P(x) dx}$.

Multiply the entire equation by $\mu(x)$ and integrate both sides.

Example: $\frac{dy}{dx} + \frac{2}{x}y = x^2$

Integrating factor: $\mu(x) = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = e^{\ln|x^2|} = x^2$

$x^2 \frac{dy}{dx} + 2xy = x^2(x^2)$

$\frac{d}{dx}(x^2 y) = x^4$

$\int \frac{d}{dx}(x^2 y) dx = \int x^4 dx$

$x^2 y = \frac{x^5}{5} + C$

$y = \frac{x^3}{5} + \frac{C}{x^2}$

Exact Equations

Equations of the form $M(x, y) dx + N(x, y) dy = 0$, where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Solve by finding a function $F(x, y)$ such that $\frac{\partial F}{\partial x} = M(x, y)$ and $\frac{\partial F}{\partial y} = N(x, y)$.

Example: $(2xy + \cos(x)) dx + (x^2 + 1) dy = 0$

$\frac{\partial M}{\partial y} = 2x$, $\frac{\partial N}{\partial x} = 2x$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.

$\frac{\partial F}{\partial x} = 2xy + \cos(x) \implies F(x, y) = xy^2 + \sin(x) + C$

$\frac{\partial F}{\partial y} = 2xy \implies N(x, y) = 2xy$, which matches the given equation.

$F(x, y) = C \implies xy^2 + \sin(x) = C$

Topic	Key Concepts	Techniques
Integration	Substitution, Integration by Parts, Trigonometric Identities, Exponential/Logarithmic Functions	Substitution, Integration by Parts, Trigonometric Identities, Standard Integrals
Volumes of Revolution	Disk/Washer Method, Shell Method	$\pi \int f(x)^2 dx$, $2\pi \int x f(x) dx$
Differential Equations	Separable Equations, Linear Equations, Exact Equations	Separation of Variables, Integrating Factor, Partial Derivatives