The substitution Rule

We use the substitution rule, a.k.a. change of variables, to deal with integration of the form

There are three major types:

$u$ is in a function

Example 1. $\int\sqrt{2x+1}dx$

Example 2. $\int 2^xdx = \int e^{ln 2 x}dx$

Example 3. $\int x\sqrt(1+x^2)dx$

Example 4. $\int x^2\sin(x^3)dx$

Example 5. $\int x^3\sqrt(1+x^2)dx$

$u$ is a denomiator

Example 6. $\int \frac{3x^2}{x^3+1}dx$

Example 7. $\int \frac{x^5}{x^3+1}dx$

Example 8. $\int \tan(x) dx =\int \frac{\sin(x)}{\cos(x)}dx$

$u$ is a numerator

Example 9. $\int\frac{\ln(x)}{x}dx$

Integration by part

The integration by part is a consequent of the multiplication rule of derivative and the fundamental theorem of Calculus. Using $d(uv) = udv + vdu$, we derive \(\int udv = uv-\int vdu.\)

The followings are examples:

Example 1 $\int \ln(x) dx$

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

Example 3 $\int x^2e^x dx$

Example 4 $\int e^x\sin(x) dx$

Example 5 $\int tan^{-1}(x)dx$ and $\int\sin^{-1}(x)dx$.

Example 6 $\int \sin^2(x)dx$

If an integral is of the form of a multiplication of two different types of functions, then we can almost certain we will use the integration by part as our first step. However, it is the only situation we will consider the integration by part. For example, the Example 6 above is not a multiplication of two different types of functions.

Use geometry to solve a definite integral

Example 10. $\int_-1^1\sqrt{1-t^2}dt$ is the area of upper semicircle.

Theorem. Let $f$ be a continuous function on $[-a,a]$. Then,

1. If $f$ is even, i.e. $f(-x)=f(x)$, then $\int_{-a}^{a}fdx = 2\int_0^afdx$.
2. If $f$ is odd, i.e. $f(-x)=-f(x)$, then $\int_{-a}^{a}fdx = 0$.

Reminder of an application of the Fundamental theorem of Calculus

Let me use another way to explain how we use the fundamental theorem of Calculus to solve $\frac{d}{dx}\int_{f(x)}^{g(x)} h(t)dt$ where $h(t)$ is a continuous function, and $f$ and $g$ are differentiable function. Let $H(x)$ is an antiderivative of $h(x)$. Then, by the fundamental theorem of Calculus, we know

Thus, its derivative is

Since $H$ is an antiderivative of $h$, we then can write the above derivative as $(h\circ g)g’ - (h\circ f)f’$.

1. If $f$ is odd, i.e. $f(-x)=-f(x)$, then $\int_{-a}^{a}fdx = 0$.