7.4 | Algebraic Identities and Trigonometric Substitutions


Trigonometric Integrals

For trigonometric integrals of the form
$$
\int \sin ^m x \cdot \cos ^n x \; dx,
$$
if

  • \(m\) is odd, then retain a factor of \(\sin x\) and use the identity \(\sin^2 x = 1- \cos^2 x\) to rewrite the remaining power of \(\sin x\) in terms of \(\cos x\)
  • \(n\) is odd, then retain a factor of \(\cos x\) and use the identity \(\cos^2 = 1 – \sin^2 x \) to rewrite the remaining power of \(\cos x\) in terms of \(\sin x\)
  • \(m\) and \(n\) are both even, then apply the identities
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2} \text{ and } \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2}
    $$
    and
    $$
    \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2}
    $$
    and
    $$
    \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$

    to reduce the powers.

Then apply \(u\)-substitution.

For trigonometric integrals of the form
$$
\int \sec^m x \cdot \tan^n x \; dx,
$$
if

  • \(n\) is odd, then retain a factor of \(\sec x \tan x\) and use the identity \(\tan^2 x = \sec^2x -1 \) to rewrite the remaining power of \(\tan x\) in terms of \( \sec x\).
  • \(m\) is even, then retain a factor of \(\sec^2 x\) and use the identity \(\sec^2x = \tan^2 x +1 \) to rewrite the remaining power of \(\sec x\) in terms of \(\tan x\).

Then apply \(u\)-substitution.

Find
$$
\int \sin^7 x \cos^4 x \; dx.
$$


$$
\begin{aligned}
\int \sin^7 x \cos^4 x \; dx & = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C \\
\end{aligned}
$$
$$
\begin{aligned}
\int & \sin^7 x \cos^4 x \; dx \\
& = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C
\end{aligned}
$$
$$
\begin{aligned}
& \int \sin^7 x \cos^4 x \; dx \\
& = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} \\
& \qquad -\frac{\cos^{11}x}{11}+C \\
\end{aligned}
$$

Trigonometric Substitution

For integrals involving Use the substitution Use the identity
\(\sqrt{a^2-x^2}, \; a>0 \) \(x = a \sin \theta, \; \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} \) \( 1- \sin^2 \theta = \cos^2 \theta \)
\(\sqrt{a^2+x^2}, \; a>0 \) \(x = a \tan \theta, \; \frac{-\pi}{2} < \theta < \frac{\pi}{2} \) \( 1+\tan^2 \theta = \sec^2 \theta \)
\(\sqrt{x^2-a^2}, \; a>0 \) \(x = a \sec \theta, \; 0 \leq \theta < \frac{\pi}{2} \text{ or } \frac{\pi}{2} < \theta \leq \pi \) \( \sec^2 \theta – 1 = \tan^2 \theta \)

Trigonometric Substitution

For integrals involving Use the substitution Use the identity
\(\sqrt{a^2-x^2}, \; a>0 \) \(x = a \sin \theta, \; \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} \) \( 1- \sin^2 \theta = \cos^2 \theta \)
\(\sqrt{a^2+x^2}, \; a>0 \) \(x = a \tan \theta, \; \frac{-\pi}{2} < \theta < \frac{\pi}{2} \) \( 1+\tan^2 \theta = \sec^2 \theta \)
\(\sqrt{x^2-a^2}, \; a>0 \) \(x = a \sec \theta, \; 0 \leq \theta < \frac{\pi}{2} \text{ or } \frac{\pi}{2} < \theta \leq \pi \) \( \sec^2 \theta – 1 = \tan^2 \theta \)

Trigonometric Substitution

  • For integrals involving \(\sqrt{a^2-x^2}, \; a>0 \)
    • use the subsitution
      $$x = a \sin \theta, $$
      where \(\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}\)
    • and the identity
      $$ 1- \sin^2 \theta = \cos^2 \theta $$
  • For integrals involving \(\sqrt{a^2+x^2}, \; a>0 \)
    • use the substitution
      $$x = a \tan \theta,$$
      where \(\frac{-\pi}{2} < \theta < \frac{\pi}{2}\)
    • and the identity
      $$ 1+\tan^2 \theta = \sec^2 \theta $$
  • For integrals involving \(\sqrt{x^2-a^2}, \; a>0 \)
    • use the substitution
      $$x = a \sec \theta,$$
      where \(0 \leq \theta < \frac{\pi}{2}\) or \(\frac{\pi}{2} < \theta \leq \pi \)
    • and the identity
      $$ \sec^2 \theta – 1 = \tan^2 \theta $$
Find
$$
\int \frac{1}{(4+x^2)^2} \; dx
$$


Let \(x = 2\tan \theta \) then \( dx = 2\sec^2 \theta \; d \theta \) and

$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$
$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$
$$
\begin{aligned}
\int & \frac{1}{(4+x^2)^2} \; dx \\
& = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$

Since \(x = 2 \tan \theta\), \( \tan \theta = \frac{x}{2} \) and \( \theta = \tan^{-1}\left( \frac{x}{2} \right) \). Also
$$
\sin \theta = \frac{x}{\sqrt{4+x^2}} \text{ and } \cos \theta = \frac{2}{\sqrt{4+x^2}}.
$$
So,

$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
\int & \frac{1}{(4+x^2)^2} \; dx \\
& = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{1}{(4+x^2)^2} \; dx \\
& = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$

Partial Fractions

A rational function
$$
\frac{P(x)}{Q(x)}
$$
with the degree of \(P(x)\) less than the degree of \(Q(x)\), can be rewritten as a sum of fractions as follows:

  • For each factor of \(Q(x)\) of the form \((ax+b)^m\), introduce terms
    $$
    \frac{A_1}{ax +b} + \frac{A_2}{(ax+b)^2}+ \cdots +\frac{A_m}{(ax+b)^m}
    $$
    $$
    \frac{A_1}{ax +b} + \frac{A_2}{(ax+b)^2}+ \cdots +\frac{A_m}{(ax+b)^m}
    $$
    $$
    \begin{aligned}
    & \frac{A_1}{ax +b} + \frac{A_2}{(ax+b)^2} \\
    & \qquad+ \cdots +\frac{A_m}{(ax+b)^m} \end{aligned}
    $$
  • For each factor of \(Q(x)\) of the form \((ax^2+bx+c)^m\), introduce terms
    $$
    \frac{A_1 x + B_1}{ax^2+bx+c}+\frac{A_1 x + B_1}{(ax^2+bx+c)^2}+ \cdots +\frac{A_1 x + B_1}{(ax^2+bx+c)^m}
    $$
    $$
    \begin{aligned}
    & \frac{A_1 x + B_1}{ax^2+bx+c}+\frac{A_1 x + B_1}{(ax^2+bx+c)^2} \\
    &+ \cdots +\frac{A_1 x + B_1}{(ax^2+bx+c)^m}
    \end{aligned}
    $$
    $$
    \begin{aligned}
    & \frac{A_1 x + B_1}{ax^2+bx+c}+\frac{A_1 x + B_1}{(ax^2+bx+c)^2} \\
    &+ \cdots +\frac{A_1 x + B_1}{(ax^2+bx+c)^m}
    \end{aligned}
    $$

where \(A_i\)’s and \(B_i\)’s are constants. At times, for ease of notation, we will use \(A,B,C,D,…\) in place of \(A_1, B_1,\) etc… If the degree of \(P(x)\) is not less than the degree of \(Q(x)\), long divide first.

Find
$$
\int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54}
$$
Hint: \(x^2+9\) is a factor of the denominator.


Since the degree of the numerator is less than the degree the denominator, we start by factoring the denominator completely. Long division of \(x^4-5x^3+15x^2-45x+54\) by \(x^2 + 9\) yields:
$$ x^2-5x+6$$
which factors as \((x-3)(x-2)\). So,

$$
\frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} = \frac{6x^3-20x^2+53x-87}{(x-3)(x-2)(x^2+9)}
$$
$$
\begin{aligned}
& \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \\
& \qquad = \frac{6x^3-20x^2+53x-87}{(x-3)(x-2)(x^2+9)}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \\
& \qquad = \frac{6x^3-20x^2+53x-87}{(x-3)(x-2)(x^2+9)}
\end{aligned}
$$

The partial fractions decomposition of the integrand will be

$$
\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9},
$$

where the constants \(A,B,C,\) and \(D\) satisfy

$$
\frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} = \frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9}.
$$
$$
\begin{aligned}
& \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \\
& \qquad = \frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9}.
\end{aligned}
$$
$$
\begin{aligned}
& \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \\
& \qquad = \frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9}.
\end{aligned}
$$

By clearing the denominator we obtain:

$$ \begin{aligned}6 & x^3 -20x^2+53x-87 \\
& = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \qquad + (Cx+D)(x-2)(x-3) \end{aligned}$$

Evaluate this equation at \(x = 1,2, 3,\) and \(4\):

$$
\begin{array}{c | c }
x & \begin{aligned}6 & x^3 -20x^2+53x-87 \\
& = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \; \; + (Cx+D)(x-2)(x-3) \end{aligned}\\ \hline
1 & \begin{aligned} & -48 = -10 A – 20 B + 2 C + 2D \\ \Rightarrow & -48 = -30 – 20 + 2C + 2D \\ \Rightarrow & 1 = C+D \end{aligned} \\ \hline
2 & -13 = -13B \Rightarrow B = 1 \\ \hline
3 & 54 = 18A \Rightarrow A = 3 \\ \hline
4 & \begin{aligned} & 189 = 50 A + 25 B + 8C + 2D \\ \Rightarrow & 189 = 150 + 25 + 8C +2D \\ \Rightarrow & 7= 4C+ D \end{aligned} \\
\end{array}
$$
$$
\begin{array}{c | c }
x & \begin{aligned}6 & x^3 -20x^2+53x-87 \\
& = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \; \; + (Cx+D)(x-2)(x-3) \end{aligned}\\ \hline
1 & \begin{aligned} & -48 = -10 A – 20 B + 2 C + 2D \\ \Rightarrow & -48 = -30 – 20 + 2C + 2D \\ \Rightarrow & 1 = C+D \end{aligned} \\ \hline
2 & -13 = -13B \Rightarrow B = 1 \\ \hline
3 & 54 = 18A \Rightarrow A = 3 \\ \hline
4 & \begin{aligned} & 189 = 50 A + 25 B + 8C + 2D \\ \Rightarrow & 189 = 150 + 25 + 8C +2D \\ \Rightarrow & 7= 4C+ D \end{aligned} \\
\end{array}
$$
$$
\begin{array}{|c | c |}
\hline
\scriptstyle x & \begin{aligned}& \scriptstyle 6 x^3 -20x^2+53x-87 \\
& \scriptstyle \quad = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \scriptstyle \qquad + (Cx+D)(x-2)(x-3) \end{aligned}\\ \hline
1 & \begin{aligned} & -48 = -10 A – 20 B + 2 C + 2D \\ \Rightarrow & -48 = -30 – 20 + 2C + 2D \\ \Rightarrow & 1 = C+D \end{aligned} \\ \hline
2 & -13 = -13B \Rightarrow B = 1 \\ \hline
3 & 54 = 18A \Rightarrow A = 3 \\ \hline
4 & \begin{aligned} & 189 = 50 A + 25 B + 8C + 2D \\ \Rightarrow & 189 = 150 + 25 + 8C +2D \\ \Rightarrow & 7= 4C+ D \end{aligned} \\ \hline
\end{array}
$$

So, \(A = 3\) and \(B = 1\). To find \(C\) and \(D\), solve the system:
$$
\begin{cases}
1 & = C &+ D \\
7 & = 4C &+ D
\end{cases}
$$

This system has solutions \(C = 2\) and \(D = -1\).

So,

$$
\begin{aligned}
& \int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \; dx \\
&= \int \left(\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9} \right) \; dx\\
&= \int \left(\frac{3}{x-3}+\frac{1}{x-2} + \frac{2x-1}{x^2+9} \right) \; dx\\
&= \int \frac{3}{x-3} \; dx + \int \frac{1}{x-2} \; dx + \int \frac{2x-1}{x^2+9} \; dx \\
& = 3\ln|x-3| + \ln|x-2| + \int \frac{2x}{x^2+9} \; dx \\
& \qquad- \int \frac{1}{x^2+9} \; dx \\
& = 3 \ln|x-3| + \ln|x-2| + \ln|x^2+9| \\
& \qquad – \frac{1}{3}\tan^{-1}\left(\frac{x}{3} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \; dx \\
&= \int \left(\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9} \right) \; dx\\
&= \int \left(\frac{3}{x-3}+\frac{1}{x-2} + \frac{2x-1}{x^2+9} \right) \; dx\\
&= \int \frac{3}{x-3} \; dx + \int \frac{1}{x-2} \; dx + \int \frac{2x-1}{x^2+9} \; dx \\
& = 3\ln|x-3| + \ln|x-2| + \int \frac{2x}{x^2+9} \; dx \\
& \qquad- \int \frac{1}{x^2+9} \; dx \\
& = 3 \ln|x-3| + \ln|x-2| + \ln|x^2+9| \\
& \qquad – \frac{1}{3}\tan^{-1}\left(\frac{x}{3} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \; dx \\
& = \int \left(\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9} \right) \; dx\\
& = \int \left(\frac{3}{x-3}+\frac{1}{x-2} + \frac{2x-1}{x^2+9} \right) \; dx\\
& = \int \frac{3}{x-3} \; dx + \int \frac{1}{x-2} \; dx \\
& \qquad + \int \frac{2x-1}{x^2+9} \; dx \\
& = 3\ln|x-3| + \ln|x-2| + \int \frac{2x}{x^2+9} \; dx \\
& \qquad – \int \frac{1}{x^2+9} \; dx \\
& = 3 \ln|x-3| + \ln|x-2| + \ln|x^2+9| \\
& \qquad – \frac{1}{3}\tan^{-1}\left(\frac{x}{3} \right) + C
\end{aligned}
$$

E 7.4 Exercises