This is a non-exhaustive list of integration techniques that are commonly used. You should be familiar with all three of the techniques listed here, and be able to apply them.
5.1. Substitution
These results can be proved using the chain rule and fundamental theorem of calculus.
For an indefinite integral:
(5.1)\[\int f(x) \textrm{d}x = \int f(x) \frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u+c\]
For a definite integral:
(5.2)\[\begin{equation}
\int_a^b f\mathrm{d}x= \int_{u(a)}^{u(b)}f\frac{\mathrm{d}x}{\mathrm{d}u}\mathrm{d}u
\end{equation}\]
Don’t forget to change the limits for definite integration!
The idea of using a substitution is that we convert a difficult integral \(\mathrm{d}x\) to an easier one \(\mathrm{d}u\), through careful choice of \(u\). Often we choose \(u\) so that its derivative cancels out a factor in \(f\), noting that the integrand in (5.1) can be written as:
(5.3)\[\begin{equation}
f \frac{\mathrm{d}x}{\mathrm{d}u}=f\biggr/\frac{\mathrm{d}u}{\mathrm{d}x}
\end{equation}\]
Worked example: \( \displaystyle \int x\sqrt{x^2-1}\mathrm{d}x\)
We can try the substitution \(u=x^2-1\).
This will eliminate the factor \(x\) and will simplify the expression \(\sqrt{x^2-1}\). We find that:
\(\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}=2x\) and so \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}u}=\frac{1}{2x}\), hence:
\[\int x\sqrt{x^2-1}\mathrm{d}x=\displaystyle \int\frac{1}{2}u^{1/2}\mathrm{d}u=\frac{1}{3}u^{3/2}+c=\frac{1}{3}(x^2-1)^{3/2}+c\]
This is the same result we can find by inspection.
Exercise 5.1
(a) Find \( I= \displaystyle\int_1^3\frac{1}{2x+1}\mathrm{d}x\).
(b) Find \( I= \displaystyle\int (ax+b)(cx+d)^n\mathrm{d}x\) using the substitution \(u=cx+d\)
Solution to (a)
To calculate the definite integral \( I=\displaystyle \int_1^3\frac{1}{2x+1}\mathrm{d}x\), we will use the substitution \(u=2x+1\). This problem could also be solved by inspection.
When \(x=1\), \(u=3\) and when \(x=3\), \(u=7\), hence, the result is
\[I=\int_{u=3}^{u=7}\frac{1}{u}\biggr/\frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}u =\frac{1}{2} \int_3^7\frac{1}{u}\mathrm{d}u=\frac{1}{2}\biggr[\ln(u)\biggr]_3^7=\frac{1}{2}\ln\biggr(\frac{7}{3}\biggr)\]
Solution to (b)
To calculate an expression of the form \( I=\displaystyle \int (ax+b)(cx+d)^n\mathrm{d}x\) we can make the substitution \(u=cx+d\), which gives:
\[I=\int\left(a\frac{u-d}{c}+b\right)u^n\frac{1}{c}\mathrm{d}u=\frac{1}{c^2}e \int(au+bc-ad)u^n\mathrm{d}u\]
By polynomial expansion:
\[I=\frac{1}{c^2}\int(a u^{n+1}+(bc-ad)u^n)\mathrm{d}u=\frac{1}{c^2}\left[\frac{a u^{n+2}}{n+2}+\frac{bc-ad}{n+1}u^{n+1}\right]\]
Rewriting in terms of \(x\) finally gives:
\[I = \frac{(cx+d)^{n+1}}{c^2}\left[\frac{a(cx+d)}{n+2}+\frac{bc-ad}{n+1}\right]\]
5.2. By parts
The following definitions can be proved using the product rule and fundamental theorem of calculus.
For an indefinite integral:
\[\int u\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = u\,v- \int v\,\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x+c\]
For a definite integral:
\[\int_a^b u\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x=\biggr[u\,v\biggr]_a^b- \int_a^b v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x\]
Proof
According to the product rule:
\[\begin{split}\mathrm{d}(u\,v) = u\,\mathrm{d}\,v + v\,\mathrm{d}u \qquad &\implies
\int u\,\mathrm{d}\,v = \int \mathrm{d}(u\,v) - \int v\,\mathrm{d}u \\
&\implies \int u\,\mathrm{d}\,v = u\,v - \int v\,\mathrm{d}u\end{split}\]
For a handy rule of thumb for choosing which term to differentiate when applying this method, we can use LIATE method:
Logarithmic Term
Inverse Term
Algebraic Term
Trignometric Term
Exponential Term
This list tells the order of preference for the term to be differentiated \(u(x) \rightarrow u'(x)\). The exponential function is at the bottom of the list because it is usually the easiest to integrate. The logarithmic term is at the top beccause it is usually the hardest.
Worked example: \(\displaystyle I = \int x \ln(x)\,\mathrm{d}x \)
We will use integration by parts to find the following integral:
Since L falls before A in LIATE, we pick:
\[\begin{split}u = \ln(x),\quad & v' = x\\
u' = \frac{1}{x}, \quad & v = \frac{1}{2}x^2\end{split}\]
and so integrating by parts we find:
\[I = \frac{1}{2} \,x^2 \,\ln(x) - \int \frac{1}{2} \,x \,\mathrm{d}x = \frac{1}{2} \, x^2 \left(\ln(x) - \frac{1}{2} \right) + C\]
Worked example: \(\displaystyle I = \int \ln(x)\,\mathrm{d}x \)
Here, we note that we can rewrite the integral as
\[I = \int x^0\,\ln(x)\,\mathrm{d}x \]
Since L falls before A in LIATE, we pick:
\[\begin{split}u = \ln(x),\quad & v' = x^0 \\
u' = \frac{1}{x},\quad &v = x^1\end{split}\]
and so integrating by parts we find:
\[I = x \ln(x) - \int \frac{x}{x}\,\mathrm{d}x = x\ln(x) - \int x^0\,\mathrm{d}x = x\ln(x) - x + C \]
Exercise 5.2
Integrate the following problems by parts:
(a) \(\displaystyle \int x e^{-x}\mathrm{d}x\qquad \qquad \)
(b) \(\displaystyle \int x^2 e^{-x}\mathrm{d}x\)
Solution to (a)
Let \(u=x,\ v'=e^{-x}\) then:
\(u'=1\), which is simpler and \(v=-e^{-x}\) which is no more complicated, so:
\[\int x e^{-x}\mathrm{d}x = -x e^{-x}+ \int e^{-x}\mathrm{d}x = -e^{-x}(x+1)+C\]
Solution to (b)
Let \(u=x^2,\, v'=e^{-x}\), then:
\(u'=2x\), which is simpler and \(v=-e^{-x}\), which is no more complicated, so:
\[\int x^2 e^{-x}\mathrm{d}x=-x^2 e^{-x}+2\int x e^{-x}\mathrm{d}x\]
From the previous example, we know the result of this integrand is:
\[\int x e^{-x}\mathrm{d}x = -e^{-x}(x+1)+C\]
and therefore we find:
\[ \int x^2 e^{-x}\mathrm{d}x=-e^{-x}(x^2+2x+2)\]
5.3. Reduction formulae
Reduction formulae set up a recurrence relation between different integrands:
\[I_{n} = f(n,x) + I_{n-a} \]
where \(a,\,n\) are integers typically and \(a < n\). Usually we solve these problems by integration by parts.
Worked example: \(\displaystyle I_n = \int_0^\infty x^n\,e^{-x}\,\mathrm{d} x\)
We can integrate by parts (using LIATE if necessary):
\[\begin{split}\begin{array}{lcl}
u(x) = x^n && v'(x) = e^{-x} \\
u'(x) = nx^{n-1} && v(x) = -e^{-x}
\end{array}\end{split}\]
Which means that we can write:
\[\begin{split}I_n &= \bigg[-x^n\,e^{-x}\bigg]_0^\infty + n\int_0^\infty x^{n-1}\, e^{-x}\,\mathrm{d} x \\
I_n &= 0 + n I_{n-1}\end{split}\]
If we were to repeat the integration by parts repeatedly we would find:
\[I_n = nI_{n-1} = n(n-1)\,I_{n-2} = \dots = n(n-1)(n-2)\dots (2)(1)\,I_0 = n!\, I_0\]
\(I_0\) is a good place to stop, since we can evaluate it:
\[I_0 = \int_0^\infty e^{-x}\,\mathrm{d} x = \bigg[-e^{-x}\bigg]_0^\infty = 0 - (-1) = 1\]
Hence \(I_n = n!\).
Solution to (a)
\[I_n = \int x^n e^x\mathrm{d}x \]
Integrating by parts:
\[\begin{split}\begin{array}{lcl}
u(x) = x^n && v'(x) = e^x \\
u'(x) = nx^{n-1} && v(x) = e^x
\end{array}\end{split}\]
and therefore:
\[I_n = \int x^n e^x\mathrm{d}x = x^n e^x - n \int x^{n-1}e^x\mathrm{d}x\]
Hence the reccurence relation is
\[I_n = x^n e^x - n I_{n-1}\]
Which means that:
\[I_0 = \displaystyle \int e^x\mathrm{d}x = e^x + C\]
and so \(I_4\) is given by:
\[\begin{split}I_4 &= e^x(x^4- 4x^3+4(3)x^2 - 4(3)(2)x +4(3)(2)(1))+C\\
&= e^x(x^4- 4x^3+12x^2 - 24x +24)+C\end{split}\]
Solution to (b)
\[I_n = \int_0^{\pi/2}\cos^{n}(x)\mathrm{d}x = \int_0^{\pi/2}\cos(x)\cos^{n-1}(x)\mathrm{d}x\]
Integrating by parts:
\[\begin{split}\begin{array}{lcl}
u(x)= \cos^{n-1}(x) && v'(x) = \cos(x) \\
u'(x) = -(n-1)\cos^{n-2}(x)\sin(x) && v(x) = \sin(x)
\end{array}\end{split}\]
and therefore:
\[\begin{split}I_n &= \left[\sin(x)\cos^{n-1}(x)\right]_0^{\pi/2} + (n-1) \int_0^{\pi/2}\sin(x)\cos^{n-2}(x)\sin(x)\mathrm{d}x \\
&=(0-0)+(n-1)\displaystyle \int\sin^2\cos^{n-2}(x)\mathrm{d}x \\
&=(n-1)\displaystyle \int(1-\cos^2(x)\cos^{n-2}(x)\mathrm{d}x \\
&=(n-1)(I_{n-2}-I_n)\end{split}\]
Rearranging gives:
\[I_n = \frac{n-1}{n}I_{n-2}\]
and so for \(I_{10}\):
\[I_{10} = \frac{9}{10}I_8 = \frac{9}{10}\frac{7}{8}I_6 = \dots = \frac{9\times7\times5\times3\times1}{10\times8\times6\times4\times2}I_0\]
Finally, noting the result for \(I_0\) gives the result for \(I_{10}\):
\[I_0 = \int_0^{\pi/2}\mathrm{d}x = \frac{\pi}{2} \qquad \implies \qquad I_{10} = \frac{63\pi}{256}\]
Solution to (c)
There may be a temptation to try to expand out the bracket or integrate by parts straight away, however the easier
method is to break up the integrand into \((1-x^3)^{n-1}\,(1-x^3)\):
\[\begin{split}I &= \int_0^1 (1-x^3)^{n-1}\,\mathrm{d} x -\int_0^1 x^3\,(1-x^3)^{n-1}\,\mathrm{d} x \\
&= I_{n-1} - \int_0^1 x\,x^2\,(1-x^3)^{n-1}\,\mathrm{d} x \\
&= I_{n-1} - \int_0^1 x\,\left(-\frac{1}{3}\right)(-3x^2)\,(1-x^3)^{n-1}\,\mathrm{d} x \end{split}\]
and then integrate the second integrand by parts:
\[\begin{split}\begin{array}{lcl}
u(x) = x && v'(x) = -\frac{1}{3} (-3x^2)(1-x^3)^{n-1}\\
u'(x) = 1 && v(x) = -\frac{1}{3n} (1-x^3)^{n}
\end{array}\end{split}\]
where we integrated \(f'(x)\) using the reverse chain rule.
\[\begin{split}\Rightarrow I_n &= I_{n-1} - \bigg[ -\frac{x}{3n} (1-x^3)^{n}\bigg]_0^1 - \frac{1}{3n}\int_0^1 (1-x^3)^{n}\,\mathrm{d} x \\
I_n &= I_{n-1} + 0 - \frac{1}{3n}I_{n} \end{split}\]
Rearranging the expression gives
\[\begin{equation*}
I_n = \frac{3n}{3n+1}I_{n-1}
\end{equation*}\]
Applying the formula recursively gives
\[\begin{split}I_4 &= \bigg(\frac{12}{13}\bigg)I_3 = \bigg(\frac{12}{13}\bigg)\bigg(\frac{9}{10}\bigg)I_2 \\
&= \bigg(\frac{12}{13}\bigg)\bigg(\frac{9}{10}\bigg)\bigg(\frac{6}{7}\bigg)I_1 =
\bigg(\frac{12}{13}\bigg)\bigg(\frac{9}{10}\bigg)\bigg(\frac{6}{7}\bigg)\bigg(\frac{3}{4}\bigg)I_0\end{split}\]
Finally, noting the result for \(I_0\) gives the result for \(I_4\):
\[\begin{equation*}
I_0 = \int_0^1\mathrm{d} x = 1 \qquad
\implies \qquad I_4 = \frac{243}{455}
\end{equation*}\]
Solution to (d)
Firstly break up the integrand into \(\sin(x) \sin^{n-1}(x)\) and then integrate by parts:
\[\begin{split}\begin{array}{lcl}
u(x) = \sin^{n-1}(x) && v'(x) = \sin(x) \\
u'(x) = (n-1)\cos(x)\sin^{n-2}(x) && v(x) = -\cos(x)
\end{array}\end{split}\]
and therefore:
\[\begin{split}I_n &= \bigg[-\cos(x)\sin^{n-1}(x)\bigg]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \cos^2(x) \sin^{n-2}(x)\,\mathrm{d} x \\
&= 0 + (n-1)\int_0^{\pi/2} (1 - \sin^2(x))\sin^{n-2}(x)\,\mathrm{d} x \\
&= (n-1)\int_0^{\pi/2} \bigg(\sin^{n-2}(x) - \sin^{n}(x)\bigg)\,\mathrm{d} x \\
I_n &= (n-1)I_{n-2} - (n-1)I_{n}\end{split}\]
Rearranging the expression gives
\[\begin{equation*}
I_n = \frac{n-1}{n}I_{n-2}
\end{equation*}\]
Applying the formula recursively gives
\[\begin{align*}
I_2 &= \frac{1}{2}I_0 = \frac{1}{2}\bigg[x\bigg]_0^{\pi/2} = \frac{\pi}{4}\\
I_3 &= \frac{2}{3}I_1 = \left(\frac{2}{3}\right)\bigg[-\cos(x)\bigg]_0^{\pi/2} = \frac{2}{3}
\end{align*}\]
notice we evaluate \(I_0\) and \(I_1\) depending on whether \(n\) is odd or even.
