Chapter 16 Further Integration Techniques

16.1 Integration by Substitution

Theorem 16.1 (Integration by Substitution) If $ u=u(x)$ is a differentiable function and \(f\) is continuous, then \[ \int f( u(x) ) \frac{du(x)}{dx} dx = \int f( u ) du \quad \text{indefinite integral} \] \[ \int_a^b f( u(x) ) \frac{du(x)}{dx} dx = \int_{u(a)}^{u(b)} f(u) du \quad \text{definite integral}. \]

Here \(d u\) acts as if it is a differential: \[ du= \frac{du(x)}{dx} dx, \;\; d (c+x)=dx; \;\; d (x^2)=2x dx, \;\; d (\sin x)= \cos x d x, ..., \]

Example 16.1 (Integration by Substitution)

  1. Evaluate the indefinite integral \[ \int (3x+2)^4 dx \] Introduce \(u=3x+2\), then \(du= \frac{du}{dx}\; dx = 3\; dx\), so \(dx=\frac{1}{3} du\). Now we can write \[\begin{align*} \int (3x+2)^4 dx&= \int u^4 \frac{du}{3}\\ &=\frac{1}{3}\frac{u^5}{5}+C\\ &=\frac{1}{15}(3x+2)^5+C. \end{align*}\]

  2. Evaluate the indefinite integral \[\int \cos(5x) dx\] Introduce \(u=5x\), then \(du = \frac{du}{dx}\; dx = 5\; dx\), so \(dx = \frac{1}{5} du\). Now we can write \[\begin{align*} \int \cos(5x) dx&=\int \cos(u)\frac{du}{5}\\ &=\frac{1}{5}\sin(u)+C\\ &=\frac{1}{5}\sin(5x)+C \end{align*}\]

  3. Evaluate the indefinite integral \[ \int 2 x \sqrt{1+x^2} dx. \] Introduce \(u=1+x^2\), then \(d u = \frac{du}{dx}\; d x=2x \; dx\) and we can write \[\begin{align*} \int \sqrt{(1+x^2)} (2x dx)&=\int {u}^{1/2} du &=\frac{2}{3} u^{3/2}+C\\ &= \frac{2}{3} (1+x^2)^{3/2}+C. \end{align*}\]

  4. Evaluate the indefinite integral \[ \int x^3 \cos (2+x^4) dx. \] Introduce \(u=2+x^4\), then \(d u =\frac{du}{dx} \; d x=4x^3 \; dx\) and we can write \begin{align*} (2+x^4) (4x^3 dx) &= (u) du &= (u) +C \ &= (2+x^4) +C. $$

  5. Evaluate the definite integral \[\int_1^2 \frac{-3}{2x} \; dx.\] Introduce \(u=2x\), then \(du=\frac{du}{dx}\; dx = 2\; dx\) and we have (remembering to change the limits \(a=1\) and \(b=2\) to \(u(a)=2\) and \(u(b)=4\)) \[\begin{align*} \int_1^2 \frac{-3}{2x} \; dx&=\frac{-3}{2}\int_2^4\frac{1}{u}\; du\\ &= [-\frac{3}{2}\ln(u)]_2^4\\ &=\frac{-3}{2}(\ln(4)-\ln(2))\\ &=\frac{-3}{2}\ln(4/2)\\ &=\frac{-3}{2}\ln(2)\\ &=-1.04 \text{to 2 d.p.} \end{align*}\]

    Or alternatively, rather than changing the limits we can substitute back in for \(u\) and use the original limits: \[ [-\frac{3}{2}\ln(2x)]_1^2=-\frac{3}{2}(\ln(4)-\ln(2))=\frac{-3}{2}\ln(2)=-1.04 \text{to 2 d.p.} \]

16.2 Integrals of trigonometric and hyperbolic functions

Here we make extensive use of trigonometric and hyperbolic identities to transform the integrand into a more manageable form.

Example 16.2 (Integrals using identities)

  1. Evaluate the definite integral \[\begin{align*} \int_0^{\pi} \sin^2(x) dx &=\int_0^{\pi} \frac{1}{2} \left( 1 - \cos(2x) \right) dx\\ &= \frac{1}{2} \left[ x - \frac{1}{2}\sin(2x) \right]_0^{\pi}\\ &= \frac{\pi}{2}\\ \end{align*}\] where we have used \(\cos(2\theta) = 1 - 2 \sin^2(\theta)\).

  2. Evaluate the indefinite integral \[\begin{align*} \int \sinh(3x)\cosh(3x) dx &= \int \frac{1}{2}\sinh(6x) dx\\ &=\frac{1}{12}\cosh(6x) + C \end{align*}\] where we have used \(\sinh(2x)=2\sinh(x)\cosh(x)\).

  3. Evaluate the indefinite integral \[\begin{align*} \int (\sin^2 (x))^2 dx & = \int_0^{\pi} \frac{1}{4} \left( 1 - \cos(2x) \right)^2 dx \\ & =\int_0^{\pi} \frac{1}{4} \left( 1 - 2\cos(2x) +\cos^2(2x) \right) dx \\ & =\int_0^{\pi} \frac{1}{4} \left[ 1 - 2\cos(2x) + \frac{1}{2} \left( 1 - \cos(4x) \right) \right] dx\\ & = \frac{1}{4} \left[ \frac{3}{2} x -\sin(2x) + \frac{1}{8} \sin(4x) \right] +C \end{align*}\] where we have used \(\cos(2\theta)=2\cos^2(\theta)-1\).

  4. Evaluate the indefinite integral \[\begin{align*} \int (\sin4x \cos 5x ) dx & = \int \frac{1}{2} \left[ \sin (-x) +\sin 9x \right] dx \\ & =\frac{1}{2} \left( \cos(x) - \frac{1}{9} \cos 9x \right) +C \end{align*}\] where we have used \(\sin(\theta)\cos(\phi)=\frac{1}{2}(\sin(\theta-\phi)+\sin(\theta+\phi))\).

  5. Evaluate the indefinite integral \[\begin{equation*} \int \frac{\sqrt{9-x^2}}{x^2} dx\\ \end{equation*}\] Making the substitution \(x=3\sin(\theta), \;\; -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), \[\begin{align*} \int \frac{\sqrt{9-x^2}}{x^2} dx & =\int \frac{\sqrt{9-(3\sin(\theta))^2} }{(3\sin(\theta))^2} d(3\sin(\theta)) \\ & = \int \frac{3\cos(\theta) }{9\sin^2(\theta) }(3\cos(\theta)) d\theta\\ &= \int \frac{\cos^2(\theta) }{\sin^2(\theta) } d \theta \\ & = \int \frac{1- \sin^2(\theta) }{\sin^2(\theta) } d \theta\\ & = \int \left( \frac{1 }{\sin^2(\theta) } -1 \right) d \theta\\ & = -\cot(\theta)-\theta+C. \end{align*}\]

16.3 Integrals of rational functions using partial fractions

Recall that a function of the form \[f(x)=\frac{p(x)}{q(x)}\] where \(p\) and \(q\) are polynomials is known as a rational function.

If \(q(x)\) is a polynomial of degree \(m\) and \(p(x)\) is a polynomial of degree \(n < m\) then we can often write a rational function as \[ \frac{p(x)}{q(x)}=\frac{c_1}{x-\alpha_1}+\dotsb + \frac{c_n}{x-\alpha_n} \] where \(\alpha_i\) are the distinct roots of \(q\). We can find the coefficients \(c_1\) by multiplying through by \(q(x)\) and equating coefficients of powers of \(x\) to obtain a system of linear equations for the \(c_i\).

We know that \[ \frac{d}{dx} \ln (x-\alpha) = \frac{1}{x-\alpha} \] and by using this fact repeatedly, we can then integrate any rational function of this form — but the answer depends on the details (because \(\ln(z)\) is only defined for positive \(z\)).

Other cases:

  • If \(p\) has degree greater than \(q\), then we can first apply polynomial long division to obtain \[\frac{p(x)}{q(x)}=s(x)+\frac{r(x)}{q(x)}\] where the degree of \(r\) will be strictly less that the degree of \(q\), and we can then perform partial fractions on the remainder term \(\frac{r}{q}\).

  • If \(q(x)\) contains an irreducible quadratic factor — factors of the form \(x^2+bx+c\) that cannot be factorised into a product of linear factors \((x-\alpha)(x-\beta)\) with real coefficients — then the numerator in the partial fractions will be a linear polynomial. For example, \[\frac{x^2+1}{(x+2)(x-1)(x^2+x+1)}=\frac{c_1}{(x+2)}+\frac{c_2}{x-1}+\frac{bx+c}{x^2+x+1}.\]

  • If \(q(x)\) contains repeated roots — factors of the form \((x-\alpha)^t\) — then the first \(t\) powers of \((x-\alpha)\) will appear as denominators (although possibly with a zero numerator). For example, \[\frac{p(x)}{(x-\alpha)^t}=\frac{c_1}{x-\alpha}+\frac{c_2}{(x-\alpha)^2}\dotsb + \frac{c_t}{(x-\alpha)^t}.\]

Example 16.3 (Integration by partial fractions)

  1. Evaluate the indefinite integral \[ I=\int \frac{2x^2 -x + 4} { x(x^2+4) } dx \] we have \[ \left[ \frac{2x^2 -x + 4} { x(x^2+4)}=\frac{A} { x}+\frac{Bx+C} {(x^2+4)}, A=1, B=1, C=-1 \right] \\ \] then \[\begin{align*} I= \int \left[ \frac{1} {x} + \frac{x-1} {x^2 +4 } \right] \\ & =\int \frac{dx} {x} + \int \frac{x} {x^2 +4 } dx -\int \frac{1} {x^2 +4 } dx \\ & = \ln |x| +\frac{1}{2} \ln (x^2 +4) - \frac{1}{2} \tan^{-1}(x/2)+C \end{align*}\]

  2. Evaluate the indefinite integral (the degree of the numerator is not less than the degree of the denominator) \[ I=\int \frac{4x^2 -3x + 2} { 4x^2 -4x +3 } dx \] We have \[\frac{4x^2 -3x + 2} { 4x^2 -4x +3 }=1+\frac{x-1} { 4x^2 -4x +3 }; 4x^2 -4x +3=( 2x -1)^2 +2\] Hence \[\begin{align*} I & =\int \left[ 1+ \frac{x-1} {(2x-1)^2+2 } \right]dx \\ &= x+ \int \left[ \frac{ (1/2)(2x-1)-(1/2) } {(2x-1)^2+2 } \right] (1/2) d(2x-1) \\ &= x+\int \left(\frac{1}{4} \frac{(2x-1)} {(2x-1)^2+2 } - \frac{1}{4} \frac{1} {(2x-1)^2+2 } \right) d(2x-1)\\ &=x+ \frac{1}{8} \ln[(2x-1)^2+2] - \frac{1}{4 \sqrt{2}}\tan^{-1} \frac{2x-1}{ \sqrt{2}} +C \end{align*}\]

16.4 Integration by Parts

According to the product rule from differentiation: If \(u(x)\) and \(v(x)\) are differentiable functions, then \[ \frac{d (u(x)v(x))}{dx}=u(x) \frac{d v(x)}{dx}+v(x) \frac{d u(x)}{dx}. \]

From the Fundamental Theorem of Calculus 15.2 it follows that \[ \int \frac{d}{dx} \left( u(x)v(x) \right) dx = u(x)v(x)= \int \left[u(x) \frac{d v(x)}{dx}+v(x) \frac{d u(x)}{dx}\right] dx, \] or \[ u(x)v(x)= \int u(x) \frac{d v(x)}{dx} dx + \int v(x) \frac{d u(x)}{dx} dx. \] Rearranging gives: \[ \int u(x) \frac{d v(x)}{dx} dx= u(x)v(x) - \int v(x) \frac{d u(x)}{dx} dx. \] This is the formula for integration by parts: \[ \int u \; dv = uv - \int v \; d u. \]

Example 16.4 (Integration by parts)

  1. Evaluate the indefinite integral \[ \int \ln x \; dx, \] where we take \[ u=\ln x, \;\; v=x \] \[\begin{align*} uv - \int v \; d u &= x \ln x - \int x d \ln x \\ & = x \ln x - \int x \left( \frac{1}{x} \; d x \right) \\ & = x \ln x - \int \; d x \\ & = x \ln x - x +C \end{align*}\]

  2. Evaluate the indefinite integral \[ \int x^2 e^x \; dx, \] where we take \[ u=x^2, \;\; v=e^x, \;\; dv=d e^x= e^x \; dx \]

    \[\begin{align*} uv - \int v \; d u&= x^2 e^x - \int e^x d x^2\\ &= x^2 e^x - \int e^x (2x d x) \end{align*}\] We need to use integration by parts again by taking \[ u=x, \;\; v=e^x, \;\; dv=d e^x= e^x \; dx \] \[\begin{align*} x^2 e^x - \left( 2 \int x d e^x \right)& = x^2 e^x - \left( 2 x e^x - 2 \int e^x d x \right) \\ & = x^2 e^x - \left( 2 x e^x - 2 e^x +C \right) \end{align*}\]

  3. Evaluate the indefinite integral \[ I=\int e^x \sin(x) \; dx, \] where we take \(u=e^x, v=-\cos(x), dv= d (-\cos(x))=\sin(x) dx\)

    \[\begin{align*} I=uv - \int v \; d u &= - e^x \cos(x) + \int \cos(x) e^x d x \; \; \\ & = - e^x \cos(x) + \left( \int e^x d \sin(x) \right) \\ & = - e^x \cos(x) + \left( \sin(x) e^x - \int \sin x d e^x\right) \\ & = - e^x \cos(x) + \sin(x) e^x - I \end{align*}\] Solving for \(I\): \[\begin{align*} I=\frac{1}{2}(- e^x \cos(x) + \sin(x) e^x) + C \end{align*}\]