3.2 Integration Formulas
3.2.1 Integration by Substitution
Let \(f\) and \(g\) be functions of \(x\), then an integral of the form \(\hspace {0.2cm}\displaystyle {\int f\big [g(x)\big ]\hspace {0.1cm}g'(x)\hspace {0.1cm}dx \hspace {0.1cm}}\) is easily evaluated by letting \(u = g(x)\), so that \(du = g'(x)dx\). Thus we get \[\int f\big [g(x)\big ]\hspace {0.1cm}g'(x)\hspace {0.1cm}dx = \int f(u)\hspace {0.1cm}du\] as long \(f\) and \(g'(x)\) are continuous
- 1.
- Find \(\hspace {0.2cm} \displaystyle {\int 3x^2\big (x^3 + 5\big )^9\hspace {0.1cm}dx}\)
- 2.
- Find \(\hspace {0.3cm} \displaystyle { \int \sin ^6x\hspace {0.1cm}\cos x \hspace {0.1cm}dx}\)
- 3.
- \(\hspace {0.3cm} \displaystyle {\int \dfrac {x}{\sqrt {1 - x^2}}\hspace {0.1cm}dx}\)
- 4.
- Find \(\hspace {0.2cm} \displaystyle {\int \dfrac {x}{\sqrt {1 - x^4}}\hspace {0.1cm}dx}\)
Solution.
Part 1
Let \(m = x^3 + 5\hspace {0.5cm} \implies \hspace {0.5cm} dm = 3x^2 dx\)
\[\int m^9\hspace {0.1cm}dm = \dfrac {m^{10}}{10} + c\hspace {1cm} \text {but}\hspace {0.3cm} m = x^3 + 5\]
\[\therefore \hspace {0.5cm} \int 3x^2\big (x^3 + 5\big )^9dx = \dfrac {\big (x^3 + 5\big )^{10}}{10} + c\]
Part 2
Let \(\hspace {0.3cm} t = \sin x \hspace {0.5cm} \implies \hspace {0.5cm} dt = \cos x\hspace {0.1cm}dx\)
\begin {align*} \implies \hspace {0.5cm}\int \sin ^6x\hspace {0.1cm}\cos x \hspace {0.1cm}dx & = \int t^2 \hspace {0.1cm}dt = \frac {t^7}{7} + c\\\\ & = \dfrac {1}{7}\sin ^7x + c\\\\ \end {align*}
Part 4
Let \(\hspace {0.2cm} u = 1 - x^4\hspace {0.5cm} \implies \hspace {0.5cm} du = -4x^3\hspace {0.1cm} dx\) Can’t work since the \(x-\)terms can not be completely eliminated. So we try \(\hspace {0.2cm} u = x^2\)
\(\implies \hspace {0.5cm} du = 2xdx \hspace {0.5cm} \implies \hspace {0.5cm} \dfrac {du}{2} = xdx\)
\begin {align*} \int \dfrac {x}{\sqrt {1 - x^4}}\hspace {0.1cm}dx & = \dfrac {1}{2}\int \dfrac {du}{\sqrt {1 - u^2}}\\\\ & = \dfrac {1}{2}\sin ^{-1} x^2 + c\\\\ \end {align*}
3.2.2 Integration by Parts
Suppose \(f\) and \(g\) are differentiable functions of \(x\). Then by the rule of derivatives of products we have \[\big (f\cdot g\big )' = f\cdot g' + f'\cdot g\] Integrating both sides with respect to \(x\), we have \begin {align*} \int \dfrac {d}{dx}\big (f(x)\big ) g(x) dx & = \int \big [ f(x)\hspace {0.1cm} g'(x) + f'(x)\hspace {0.1cm}g(x)\big ]dx\\\\ f(x)\hspace {0.1cm}g(x) & = \int f(x)\hspace {0.1cm}g'(x)\hspace {0.1cm}dx + \int f'(x)\hspace {0.1cm} g(x)\hspace {0.1cm} dx\\\\ \implies \hspace {0.5cm} \int f(x)\cdot g'(x) dx & = f(x)\cdot g(x) - \int f'(x) \cdot g(x) dx \end {align*}
This is the integration by parts formula.
This formula is easily remembered when we make the following substitution. We take
\(\hspace {0.2cm} u = f(x)\hspace {0.2cm}\) and \(\hspace {0.2cm} dv = g'(x)dx\). Then \(\hspace {0.2cm} du = f'(x)dx\hspace {0.2cm}\) and \(\hspace {0.2cm} v = \displaystyle {\int g'(x)dx}\)
Substituting in the formula yields \[\int u\hspace {0.1cm}dv = u\hspace {0.1cm}v - \int v\hspace {0.1cm}du\]
Example 3.2.2. Evaluate the integrals
- 1.
- \(\displaystyle {\int x^2e^{2x}\hspace {0.1cm}dx}\)
- 2.
- \(\displaystyle {\int x^5 \ln x\hspace {0.1cm} dx}\)
Solution.
Part 1
Let \(\hspace {0.2cm} u = x^2\hspace {0.2cm}\) and \(\hspace {0.2cm} dv = e^{2x}.\hspace {0.2cm}\) Then \(\hspace {0.2cm} du = 2x \hspace {0.1cm}dx\hspace {0.2cm}\) and \(\hspace {0.2cm} v = \dfrac {e^{2x}}{2}.\hspace {0.2cm}\) Thus
\begin {align*} \int x^2e^{2x}\hspace {0.1cm}dx & = \int u dv = uv - \int v du\\\\ & = \frac {1}{2}e^{2x}x^2 - \dfrac {1}{2}\int 2 e^{2x}x\hspace {0.1cm}dx\\\\ & = \dfrac {1}{2}x^2e^{2x} - \int x e^{2x}\hspace {0.1cm}dx \end {align*}
To evaluate \(\hspace {0.2cm}\displaystyle {\int x e^{2x}\hspace {0.1cm}dx\hspace {0.2cm}}\) we use integration by parts again. Let \(\hspace {0.2cm} u = x\hspace {0.2cm}\) and \(\hspace {0.2cm} dv = e^{2x}\hspace {0.1cm}dx.\hspace {0.3cm}\) Then \(\hspace {0.2cm} du = dx\hspace {0.2cm}\) and \(\hspace {0.2cm} v = \dfrac {1}{2}e^{2x}\)
\begin {align*} \int x e^{2x}\hspace {0.1cm}dx & = \dfrac {1}{2}xe^{2x} - \dfrac {1}{2}\int e^{2x}\hspace {0.1cm}dx\\\\ & = \dfrac {1}{2}x e^{2x} - \dfrac {1}{4}e^{2x} + c_1\\ \end {align*}
\begin {align*} \implies \hspace {0.5cm} I & = \dfrac {1}{2}x^2 e^{2x} - \Big [ \dfrac {1}{2}x e^{2x} - \dfrac {1}{4}e^{2x} + c_1\Big ]\\\\ & = \dfrac {1}{2}e^{2x}\Big [x^2 - x + \dfrac {1}{2}\Big ] + c\\ \end {align*}
Note. If \(\hspace {0.2cm} u = e^{2x}\hspace {0.2cm}\) and \(\hspace {0.2cm} dv = x^2\hspace {0.1cm}dx\), then \(\hspace {0.2cm}du = \dfrac {1}{2}e^{2x}\hspace {0.2cm}\) and \(\hspace {0.2cm} v = \dfrac {1}{3}x^3\)
\[\implies \hspace {0.5cm} \int x^2 e^{2x} \hspace {0.1cm}dx = \dfrac {1}{3}x^3e^{2x} - \frac {1}{6}\int x^3e^{2x}\hspace {0.1cm}dx\]
This integral will not terminate because of wrong choice of \(u\) and \(dv\)
Part 2
Let \(\hspace {0.2cm} u = \ln x\hspace {0.3cm}\implies du = \dfrac {1}{x}dx\hspace {0.3cm}\) and \(\hspace {0.3cm} dv = x^5\hspace {0.1cm}dx\hspace {0.3cm}\implies \hspace {0.3cm} v = \dfrac {1}{6}x^6.\hspace {0.3cm}\) Thus
\begin {align*} \int x^5 \ln x\hspace {0.1cm} dx & = \dfrac {1}{6}x^6\ln x - \frac {1}{6}\int x^6\cdot \frac {1}{x}\hspace {0.1cm}dx\\\\ & = \frac {1}{6}x^6\ln x - \frac {1}{6}\int x^5\hspace {0.1cm}dx\\\\ & = \dfrac {1}{6}x^6\ln x - \frac {1}{36}x^6 + c\\\\ & = \frac {1}{6}x^6\Big [ \ln x - \dfrac {1}{6}\Big ] + c\\\\ \end {align*}
To determine which part of the integral is \(u\) and the other is \(dv\) in the application of the integration by parts formula, then following two rules may be used:
- 1.
- The part selected as \(dv\) must be readily integrable.
- 2.
- \(\displaystyle {\int v du}\hspace {0.3cm}\) must not be more complex than \(\displaystyle {\int u dv}\).
However, the following lists of common integrals with suggestions for the choice of \(u\) and \(dv\)
- 1.
- For integrals of the form \(\hspace {0.3cm}\displaystyle {\int x^n e^{kx}\hspace {0.1cm}dx\hspace {0.3cm}, \hspace {0.3cm} \int x^n \sin kx \hspace {0.1cm}dx}\hspace {0.3cm}\) or \(\hspace {0.3cm} \displaystyle {x^n\cos kx \hspace {0.1cm}dx}\)
Let \(\hspace {0.2cm} u = x^n\hspace {0.2cm}\) and \(\hspace {0.2cm}\displaystyle { dv = e^{kx}\hspace {0.1cm}dx}\hspace {0.2cm}\) or \(\hspace {0.2cm} \sin kx\hspace {0.1cm}dx\hspace {0.1cm}\) or \(\hspace {0.1cm} \cos kx\hspace {0.2cm}dx\)
- 2.
- For the integrals of the form \(\hspace {0.3cm}\displaystyle {\int x^n\ln x\hspace {0.1cm}dx\hspace {0.3cm} , \hspace {0.3cm} \int x^n\sin ^{-1}kx\hspace {0.1cm}dx\hspace {0.3cm}}\) or \(\hspace {0.3cm}\displaystyle {\int x^n \cos ^{-1}kx\hspace {0.1cm}dx}\hspace {0.3cm}\) or \(\hspace {0.3cm}\displaystyle {\int x^n \tan ^{-1}kx\hspace {0.1cm}dx}\)
Let \(\hspace {0.2cm} u = \ln x \hspace {0.3cm} , \hspace {0.3cm} \sin ^{-1}kx\hspace {0.3cm}\) or \(\hspace {0.3cm} \cos ^{-1}kx\hspace {0.3cm}\) or \(\hspace {0.3cm} \tan ^{-1}kx\hspace {0.3cm}\) and \(\hspace {0.3cm} dv = x^n\hspace {0.1cm}dx\)
- 3.
- For integrals of the form \(\hspace {0.2cm} \displaystyle {\int e^{ax}\sin bx\hspace {0.1cm}dx}\hspace {0.4cm} \) or \(\hspace {0.4cm}\displaystyle {\int e^{ax}\cos bx\hspace {0.1cm}dx}\)
Let \(\hspace {0.2cm} u = \sin bx\hspace {0.3cm}\) or \(\hspace {0.3cm} \cos bx\hspace {0.3cm}\) and \(\hspace {0.3cm} dv = \displaystyle {e^{ax}\hspace {0.1cm}dx}\)
Find the integral \(\hspace {0.2cm}\displaystyle {I = \int e^x\cos x \hspace {0.1cm}dx}\)
Solution.
Let \(\hspace {0.2cm} u = \cos x\hspace {0.2cm} \) and \(\hspace {0.2cm} dv = \displaystyle {e^x\hspace {0.1cm}}.\hspace {0.2cm}\) Then \(\hspace {0.2cm} du = -\sin x\hspace {0.1cm}dx\hspace {0.2cm}\) and \(dv = \displaystyle {e^x}\hspace {0.2cm}\) Thus
\[I = \int e^x\cos x\hspace {0.1cm}dx = e^x\cos x + \int e^x \sin x \hspace {0.1cm}dx\]
To evaluate \(\hspace {0.2cm}\displaystyle {\int e^x \sin x \hspace {0.1cm}dx},\hspace {0.2cm}\) we let \(\hspace {0.2cm} u = \sin x \hspace {0.2cm}\) and \(\hspace {0.2cm} dv = e^x \hspace {0.1cm}dx.\hspace {0.3cm}\) Then \(\hspace {0.2cm} du = \cos x\hspace {0.2cm}\) and \(\hspace {0.2cm} v = e^x\)
\begin {align*} \implies \hspace {0.5cm} \int e^x \sin x \hspace {0.1cm}dx & = e^x \sin x - \int e^x\cos x \hspace {0.1cm}dx\\\\ & = e^x\sin x - I \end {align*}
\begin {align*} \text {Thus}\hspace {0.5cm} I & = e^x\cos x + \Big [e^x\sin x - I\Big ]\\\\ \implies \hspace {0.5cm} 2I & = e^x\cos x + e^x\sin x\\ \end {align*}
\[\therefore \hspace {0.5cm} I = \int e^x\cos x\hspace {0.1cm}dx = \frac {1}{2}e^x\Big [\cos x + \sin x \Big ] + c\]
3.2.3 Reduction Formula
When using integration by parts you may find that it is necessary to apply the formula repeatedly. The labour involved in successive application of integration by parts may be reduce by use of the reduction formula.
For example, suppose \(\hspace {0.2cm} \displaystyle {I_n = \int x^ne^{ax}\hspace {0.2cm}dx}\). Here \(\hspace {0.2cm} u = x^n\hspace {0.2cm}\) and \(\hspace {0.2cm} dv = \displaystyle {e^{ax}\hspace {0.2cm}dx}\)
\(\implies \hspace {0.5cm} \displaystyle {du = n x^{n - 1}\hspace {0.1cm}dx}\hspace {0.5cm}\) and \(\hspace {0.5cm}\displaystyle {v = \dfrac {1}{a}e^{ax}}\)
\[\implies \hspace {0.5cm} I_n = \frac {1}{a}x^ne^{ax} - \dfrac {n}{a}\underbrace {\int x^{n-1}e^{ax}\hspace {0.1cm} dx}_{I_{n-1}}\]
\[\implies \hspace {0.5cm} I_n = \frac {1}{a}x^ne^{ax} - \dfrac {n}{a}I_{n-1}\hspace {0.5cm} n\geq 1\]
This is the reduction formula of the integral \(\hspace {0.3cm}\displaystyle {I_n = \int x^n e^{ax}\hspace {0.1cm}dx}\)
Evaluate the integral \(\hspace {0.1cm} \displaystyle {\int x^3 e^{2x}\hspace {0.1cm}dx}\)
Solution.
Let \(\hspace {0.2cm} u = x^3\hspace {0.3cm}\) and \(\hspace {0.3cm} dv = \displaystyle {e^{2x}\hspace {0.1cm}dx}\). Then \(\hspace {0.2cm} du = 3x^2dx\hspace {0.3cm}\) and \(\hspace {0.3cm} v = \displaystyle {\frac {1}{2}e^{2x}}\hspace {0.2cm}\) Thus
\[I_3 = \int x^3 e^{2x}\hspace {0.1cm}dx = \dfrac {1}{2}x^3e^{2x} - \dfrac {3}{2}\int x^2e^{2x}dx\]
\begin {align*} \implies \hspace {0.5cm} I_3 & = \dfrac {1}{2}x^3e^{2x} - \dfrac {3}{2}I_2\\\\ I_2 & = \dfrac {1}{2}x^2e^{2x} - \dfrac {2}{2}I_1\\\\ I_1 & = \dfrac {1}{2}xe^{2x} - \dfrac {1}{2}I_0\\\\ I_0 & = \int e^{2x}dx = \dfrac {1}{2}e^{2x} \end {align*}
Using back substitution, we have
\begin {align*} I_1 & = \dfrac {1}{2}xe^{2x} - \dfrac {1}{2}\cdot \dfrac {1}{2}e^{2x}\\\\ I_2 & = \dfrac {1}{2}x^2e^{2x} - \Bigg (\dfrac {1}{2}xe^{2x} - \dfrac {1}{4}e^{2x}\Bigg ) = \dfrac {1}{2}x^2e^{2x} - \dfrac {1}{2}xe^{2x} + \dfrac {1}{4}e^{2x}\\\\ \therefore \hspace {0.5cm} I_3 & = \dfrac {1}{2}x^3e^{2x} - \dfrac {3}{2}\Bigg [\dfrac {1}{2}x^2e^{2x} - \dfrac {1}{2}xe^{2x} + \dfrac {1}{4}e^{2x}\Bigg ]\\ \end {align*}
\[\therefore \hspace {0.5cm} I_3 = \dfrac {1}{2}e^{2x} \Bigg [ x^3 - \dfrac {3}{2}x^2 + \dfrac {3}{2}x - \dfrac {3}{4}\Bigg ] + c\]
3.2.4 The Method of Undetermined Coefficient
Solution.
Let \(\hspace {0.3cm} I_4 = \displaystyle {e^{2x}\hspace {0.1cm}P(x)}\), where \[P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4\] Thus, \(\hspace {0.3cm}\displaystyle {I_4 = e^{2x}\Big [a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 \Big ]}\hspace {0.2cm}\) where \(a_0,a_1, a_2, a_3\) and \(a_4\) are the coefficients
\begin {align*} \dfrac {d I_4}{dx} & = 2e^{2x}\Big [a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4\Big ] + e^{2x}\Big [a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3\Big ]\\\\ & = e^{2x}\Big [(2a_0+a_1) + (2a_1 + 2a_2) x + (2a_2 + 3a_3)x^2 + (2a_3 + 4a_4)x^3 + 2a_4x^4\Big ] = x^4e^{2x}\\ \end {align*}
\[\implies \hspace {0.3cm} 2a_0 + a_1 = 0\hspace {0.2cm}, \hspace {0.2cm} 2a_1 + 2a_2 = 0\hspace {0.2cm}, \hspace {0.2cm} 2a_2 + 3a_3 = 0\hspace {0.2cm} , \hspace {0.2cm} 2a_3 + 4a_4=0\hspace {0.3cm} \text {and}\hspace {0.3cm} 2a_4 = 1 \implies a_4 = \dfrac {1}{2}\]
\[\implies \hspace {0.3cm} a_3 = -1\hspace {0.3cm}, \hspace {0.3cm} a_2 = \dfrac {3}{2}\hspace {0.3cm} , \hspace {0.3cm} a_1 = \dfrac {-3}{2} \hspace {0.3cm}\text {and}\hspace {0.3cm} a_0 = \dfrac {3}{4}\]
\[\therefore \hspace {0.5cm} I_4 = e^{2x}\Big [\dfrac {3}{4} - \dfrac {3}{2}x + \dfrac {3}{2}x^2 - x^3 + \dfrac {1}{2}x^4\Big ] + c\]
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