Integration of substitutions
Kapitoly: Integral, Integration per partes, Integration of substitutions, The definite integral
The substitution method is a very effective method in integration. During integration, we replace part of a function with another function, integrate and substitute back.
The basic formula
Consider the function f and its primitive function F on the interval J. Next, consider the real function $\phi$, which is derivable on the interval I and yet $\phi(I) \subseteq J$. Then
$$ \int f(\phi(t))\cdot\phi^{\prime}(t) \mbox{ d}t = F(\phi(t)). $$
How did we get from the left side to the right side?
We will use this theorem by choosing the substitution $x = \phi(t)$ and changing the integration variable - from dt we get dx. So we start modifying the integral on the left-hand side by just substituting $\phi(t)$ for x and rewriting dt to dx:
$$ \int f(\phi(t))\cdot\phi^{\prime}(t) \mbox{ d}t = \int f(x) \cdot x^{\prime} \mbox{ d}x $$
Note that if we substitute $x = \phi(t)$, we suddenly have the derivative calculated (we are already deriving by x): $x^{\prime} = 1$. So we can throw out the derivative of x from the integral:
$$ \int f(x) \cdot x^{\prime} \mbox{ d}x = \int f(x) \mbox{ d}x $$
Furthermore, by assumption, F is a primitive function to the function f, so we can replace the whole integral with the function F:
$$ \int f(x) \mbox{ d}x = F(x) $$
Finally, we undo the substitution, writing the original $\phi(t)$ instead of x:
$$ F(x) = F(\phi(t)) $$
We have thus obtained the equality from the introduction:
$$ \int f(\phi(t))\cdot\phi^{\prime}(t) \mbox{ d}t = \ldots = F(\phi(t)). $$
This is approximately how we will use this sentence.
Example one
Calculate this integral:
$$ \int \sin 6t \mbox{ d}t $$
We know the formula for the integral sin x (or here sin t), but we don't know the formula for sin 6x. So we use substitution. We will substitute as follows: x = 6t. According to the previous formula:
$$ \int f(\phi(t))\cdot\phi^{\prime}(t) \mbox{ d}t = \int f(x) \mbox{ d}x, $$
where $x = \phi(t)$. For our particular case, f = sin and $\phi(t) = 6t$. Thus, according to the formula, the following relationship holds:
$$ \int (\sin 6t) \cdot (6t)^{\prime} \mbox{ d}t = \int \sin x \mbox{ d}x, $$
where x = 6t. The problem is that we want to compute the integral of sin 6t, not the integral of $(\sin 6t) \cdot (6t)^{\prime}$. How to get out of this? Let's try to derive the function 6t. We get:
$$ \int (\sin 6t) \cdot 6 \mbox{ d}t = \int \sin x \mbox{ d}x, $$
The number six is a constant that we can put in front of the integral:
$$ 6\cdot\int (\sin 6t) \mbox{ d}t = \int \sin x \mbox{ d}x, $$
Now we divide the whole equation by six:
$$ \int (\sin 6t) \mbox{ d}t = \frac16\cdot\int \sin x \mbox{ d}x, $$
On the left side we have the original function we wanted to calculate at the beginning. We just need to calculate the integral on the right side. The integral of sin x is equal to −cos x:
$$ \int (\sin 6t) \mbox{ d}t = \frac16\cdot(-\cos x), $$
And after x we add back 6t:
$$ \int (\sin 6t) \mbox{ d}t = -\frac16\cdot\cos 6t + c, $$
And that's the finished result.
Example one easier
The previous procedure was correct, but usually the substitution is written a little differently. Again the assignment:
$$ \int \sin 6t \mbox{ d}t $$
Write the substitution in parentheses after the integral as follows:
$$ \int \sin 6t \mbox{ d}t = \begin{bmatrix} x=6t\\ dx=6dt \end{bmatrix} $$
In the first line, we write our substitution, i.e. x = 6t. Now we will derive both sides. The left one by x and the right one by t. In the second row of the first column we will always have dx, because the derivative of x always comes out 1. Thus the expression dx just indicates that we have derived by x.
The right-hand side we derive normally by t. We get $(6t)^{\prime} = 6$ and add dt, because we derived by t. If we want to make our work easier, we can still write dt:
$$ \int \sin 6t \mbox{ d}t = \begin{bmatrix} x=6t\\ dx=6dt\\ \frac{dx}{6}=dt \end{bmatrix} $$
In the next step, we do the substitution itself. So what are we going to substitute for what? We want to have the expression x instead of 6t, so that will be the first substitution. Since we're inserting a new variable x into the function , we need to change dt at the same time. So instead of dt, we'll substitute the isolated value in the third line, dx/6.
Recap: instead of 6t, we write x (first line) and instead of dt, we write dx/6 (third line). We get:
$$ \int \sin 6t \mbox{ d}t = \begin{bmatrix} x=6t\\ dx=6dt \end{bmatrix} = \int \sin x \frac{dx}{6} $$
From here on, the procedure is completely normal. Put the six in front of the integral:
$$ \int \sin x \frac{dx}{6} = \frac16\cdot\int \sin x dx $$
Integrate the sine:
$$ \frac16\cdot\int \sin x dx = \frac16 \cdot (-\cos x) $$
And insert the expression 6t back after x:
$$ \frac16 \cdot (-\cos x) =
- \frac16 \cdot \cos 6t + c $$
Second example
Calculate the integral using the substitution method:
$$ \int \tan t \mbox{ d}t $$
We could probably find the tangent in some tables, but we can easily calculate this integral ourselves. The first thing to do is to decompose the tangent into the quotient sin/cos:
$$ \int \tan t \mbox{ d}t = \int \frac{\sin t}{\cos t} \mbox{ d}t $$
We choose the substitution as follows: x = cos t. We get:
$$ \int \frac{\sin t}{\cos t} \mbox{ d}t = \begin{bmatrix} x=\cos t\\ dx=-\sin t \mbox{ d}t \end{bmatrix} $$
Now let's isolate dt:
$$ dt=-\frac{dx}{\sin t} $$
Now we can substitute into the integral. Instead of cos t we write x and instead of dt we write −dx/sin t.
$$ \int \frac{\sin t}{\cos t} \mbox{ d}t = \begin{bmatrix} x=\cos t\\ dx=-\sin t \mbox{ d}t\\ \end{bmatrix} = \int -\frac{\sin t}{x}\cdot\frac{dx}{\sin t} $$
A minor problem is that we have left the previous integer variable t, in the expression, in the sin. Fortunately, though, both of our sines get wiped out and we're left with :
$$ \int -\frac{\sin t}{x}\cdot\frac{dx}{\sin t} = \int -\frac{dx}{x} $$
Let's break this down a bit to get a nicer expression out of it. We move dx after the fraction and put a one in the numerator. We'll move the minus sign in front of the integral:
$$ \int -\frac{dx}{x} = -\int \frac{1}{x} \mbox{ d}x $$
We already know the integral from 1/x, it's the logarithm. So we get:
$$ -\int \frac{1}{x} \mbox{ d}x = -\ln |x| $$
And we put the original cos t back after x.
$$ -\int \frac{1}{x} \mbox{ d}x = -\ln x = -\ln |\cos t| + c $$
The third example
Calculate the integral using the substitution method:
$$ \int x \cdot \sin^3 t \mbox{ d}t $$
You may be confused by the fact that we have two variables in the integrand: x and t. It doesn't matter, because we know from dt that we are integrating by the variable t and so x behaves like a constant. So let's not worry the wolf any more and move x in front of the integral:
$$ \int x \cdot \sin^3 t \mbox{ d}t = x\cdot\int \sin^3 t \mbox{ d}t $$
We don't have many formulas for sin3, but we could find something for sin2, so let's modify the function as follows:
$$ x\cdot\int \sin^3 t \mbox{ d}t = x\cdot\int \sin^2t\cdot\sin t \mbox{ d}t $$
Now we decompose sin2 according to the formula sin2t = (1−cos2t). We get the integral:
$$ x\cdot\int \sin^2t\cdot\sin t \mbox{ d}t = x\cdot\int (1-\cos^2t)\cdot\sin t \mbox{ d}t $$
Now we make the substitution. We have the variable x taken, so we'll use y, for example. And we choose y = cos t
$$ x\cdot\int (1-\cos^2t)\cdot\sin t \mbox{ d}t = \begin{bmatrix} y=\cos t\\ dy = -\sin t dt\\ \end{bmatrix} $$
We'll make dt stand alone:
$$ dt=-\frac{dy}{\sin t} $$
And put it in the integral:
$$ x\cdot\int (1-\cos^2t)\cdot\sin t \mbox{ d}t = \begin{bmatrix} y=\cos t\\ dy = -\sin t dt\\ \end{bmatrix}= -x\cdot\int(1-y^2)\cdot\sin t \cdot \frac{dy}{\sin t} $$
Siny will be nicely truncated again:
$$ -x\cdot\int(1-y^2)\cdot\sin t \cdot \frac{dy}{\sin t}=-x\cdot\int(1-y^2) \cdot dy $$
We can now decompose this integral into two integrals:
$$ -x\cdot\int(1-y^2) \cdot dy=-x\cdot\left(\int1 \mbox{ d}y-\int y^2 \mbox{ d}y \right) $$
These are all table values:
$$ -x\cdot\left(\int1 \mbox{ d}y-\int y^2 \mbox{ d}y \right)=-x\cdot\left(y-\frac{y^3}{3}\right) $$
Now we add back after y the original cos t:
$$ -x\cdot\left(y-\frac{y^3}{3}\right)=-x\cdot \left(\cos t-\frac{\cos^3t}{3}\right)+c $$
This is the final result.