Integration by Substitution
Integration by substitution (as seen in SL 5.10) simplifies complex integrals into a form with which we can integrate.
The general method can be seen as.
$$\int kf(g(x))g'(x)dx$$
where we make $u = g(x)$ such that $du = g'(x)dx$ and $dx = \frac{du}{g'(x)}$.
Therefore the integral becomes
$$k\int f(u)g'(x)\,\frac{du}{g'(x)} = k \int f(u)\,du$$
where $k$ is a constant.
Of course we would not want to do this if $f(x)$ is a function that's even more difficult to integrate since it is a method to simplify! Choose $u=g(x)$ wisely.
In IB examinations, if the integral is not in the form $\int kg'(x)f(g(x))dx$, the substitution will be provided.
Steps for Integration by Substitution:
- Identify a suitable—suitable based on your judgement—substitution $u = g(x)$
- Calculate $ \frac{du}{dx} = g'(x)$
- Express the integral in terms of $u$
- Integrate with respect to $u$
- Substitute back to express the result in terms of $x$
Let's integrate $\int x\sqrt{1-x^2}dx$
- Let $u = 1-x^2$
- $du = -2xdx$ or $-\frac{1}{2}du = xdx$
- The integral becomes $-\frac{1}{2}\int \sqrt{u}du$
- Integrating: $-\frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = -\frac{1}{3}u^{3/2} + C$
- Substituting back: $-\frac{1}{3}(1-x^2)^{3/2} + C$
Look for parts of the integrand that appear both as a function and its derivative. These are often good candidates for substitution.
Remember that for a definite integral in the form
$$\int^a_b f(x) dx$$
when doing the substitution $u = g(x)$, it is necessary to change the bounds of the integral since we're now integrating with respecting to $u$ and not $x$.
So the integral becomes:
$$\int^{g(a)}_{g(b)} f(g^{-1}(u))g'(g^{-1}(u)) du$$
Which will hopefully be easier than whatever you were doing before.
Integration by substitution can often be tricky because it can easily make an integrand more complicated if you choose the wrong substitution. Make sure to think about what the substitution will do before you choosing it, and don't be afraid to start over.
Do not try to memorise every substitution for every integral. They seem stupid to memorise—and they are. You'll learn these work with enough practice. Eventually you'll be able to come up with your own substitutions seemingly out of thin air.
Integration by Parts
Integration by parts is a method used when the integrand is a product of two functions. It's based on the product rule of differentiation.
Let $h(x) = f(x)g(x)$ such that we have a function of $x$ that can be written as a product of two, then by applying the product rule we get,
$$h'(x) = (f(x)g(x))' = f'(x)g(x) + g'(x)f(x)$$
Which we can rearrange to get
$$(f(x)g(x))' - f'(x)g(x) = g'(x)f(x)$$
And then integrate both sides to get
$$ f(x)g(x) - \int f'(x)g(x)\,dx = \int g'(x)f(x)\,dx$$
Hence Integration by parts can be written as follows,
$$\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$$
where we simplify the notation to $u$ and $v$ which are functions of $x$.
Steps for Integration by Parts:
- Choose $u$ and $\frac{dv}{dx}$ from the integrand
- Calculate $\frac{du}{dx}$ and $v$
- Apply the formula
- Solve the resulting integral
Let's integrate $\int x\cos(x)dx$
- Choose $u = x$ and $dv/dx = \cos(x)$ such that you select one function and one derivative
- Then derive $u$ such that $du/dx = 1$ and integrate $\frac{dv}{dx}$ to get $v = \sin(x)$
- Applying the formula: $\int x\cos(x)dx = x\sin(x) - \int \sin(x)dx$
- Solving the resulting integral: $x\sin(x) + \cos(x) + C$
Students often struggle with choosing which part of the integrand should be $u$ and which should be $\frac{dv}{dx}$. A helpful acronym is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which suggests the order of preference for choosing $u$. In other words, choose $u$ as the term that looks harder to integrate and $\frac{dv}{dx}$ as the term that is easy to integrate—differentiation is much more straightforward than integration.
Repeated Integration by Parts
Some integrals require multiple applications of integration by parts. This technique is known as repeated integration by parts.
Let's integrate $\int x^2e^x dx$
- First application: Let $u = x^2$ and $dv/dx = e^x$
Then $du/dx = 2x$ and $v = e^x$
hence $\int x^2e^x dx = x^2e^x - \int 2xe^x dx$ - Second application (on the new integral): Let $u = 2x$ and $dv/dx = e^x$
Then $du/dx = 2$ and $v = e^x$
Hence $\int 2xe^x dx = 2xe^x - \int 2e^x dx$ - Combining results: $\int x^2e^x dx = x^2e^x - (2xe^x - 2e^x) + C$
leads to $= x^2e^x - 2xe^x + 2e^x + C$
Another classic example of repeated integration by parts is $\int e^x \sin(x) dx$.
This integral is particularly interesting because it creates a cycle that eventually leads back to the original integral.
- First application: Let $u = \sin(x)$ and $dv/dx = e^x$
Then $du/dx = \cos(x)$ and $v = e^x$
hence $\int e^x\sin(x)\, dx = e^x\sin(x) - \int e^x \cos(x)\,dx$ - Second application (on the new integral): Let $u = \cos(x)$ and $dv/dx = e^x$
Then $du/dx = -\sin(x)$ and $v = e^x$
Hence $$\begin{align*} \int e^x\cos(x)\, dx &= e^x\cos(x) - \int e^x(-\sin(x))\, dx \\ &= e^x\cos(x) + \int e^x\sin(x)\,dx \end{align*}$$ - Combining results: $$\begin{align*}\int e^x\sin(x) dx &= e^x\sin(x) - \left(e^x\cos(x)+\int e^x\sin(x)\,dx \right) + C \\ &= e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)\,dx + C \end{align*}$$
- Therefore by moving $\int e^x\sin(x)\,dx$ do the other side we get $$\begin{align*} 2\int e^x\sin(x)\,dx & = e^x\sin(x) - e^x\cos(x) + C \\ & \Longrightarrow \int e^x\sin(x)\,dx = \frac{e^x\sin(x) - e^x\cos(x)}{2} + C_1 \end{align*}$$
Where $\frac{C}{2}$ is just another constant so we use $C_1$
The technique of repeated integration by parts is particularly useful for integrals involving polynomials multiplied by exponential, trigonometric, or logarithmic functions.
When dealing with repeated integration by parts, it's often helpful to organize your work in a table format to keep track of the different terms and avoid mistakes.
