Class 12 Integration: Partial Fractions and Substitution (Part 2)

This lesson solves four integration questions from Exercise 7.5, each one combining a substitution with a partial fraction split. The first two remove the trigonometry by substituting for the sine, and the last two simplify a ratio of even powers by substituting for $x^2$.

First integral: a substitution leading to a repeated factor

We want to evaluate

$I = \int \frac{(3\sin x - 2)\cos x}{5 - \cos^2 x - 4\sin x} \, dx.$

Replace $\cos^2 x$ with $1 - \sin^2 x$ in the denominator:

$5 - (1 - \sin^2 x) - 4\sin x = 4 + \sin^2 x - 4\sin x.$

Put $\sin x = y$, so $\cos x \, dx = dy$, and the integral becomes rational:

$I = \int \frac{3y - 2}{y^2 - 4y + 4} \, dy = \int \frac{3y - 2}{(y - 2)^2} \, dy.$

Partial fractions

Because $(y - 2)^2$ is a repeated factor, we write

$\frac{3y - 2}{(y - 2)^2} = \frac{A}{y - 2} + \frac{B}{(y - 2)^2}.$

Multiplying through by $(y - 2)^2$ gives $3y - 2 = A(y - 2) + B$. Putting $y = 2$ leaves $B = 3(2) - 2 = 4$, and comparing the coefficients of $y$ gives $A = 3$. So

$\frac{3y - 2}{(y - 2)^2} = \frac{3}{y - 2} + \frac{4}{(y - 2)^2}.$

Integrating

Using $\int \dfrac{dx}{x^2} = -\dfrac{1}{x}$,

$I = 3\log|y - 2| - \frac{4}{y - 2} + C.$

Now substitute $y = \sin x$. Since $2 - \sin x$ is always positive, $|y - 2| = 2 - \sin x$, and

$I = 3\log(2 - \sin x) + \frac{4}{2 - \sin x} + C.$

Question 17: a substitution giving two simple factors

We want

$I = \int \frac{\cos x}{(1 - \sin x)(2 - \sin x)} \, dx.$

Put $\sin x = t$, so $\cos x \, dx = dt$:

$I = \int \frac{dt}{(1 - t)(2 - t)}.$

Partial fractions

Write

$\frac{1}{(1 - t)(2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t},$

so $1 = A(2 - t) + B(1 - t)$. Putting $t = 1$ gives $A = 1$, and putting $t = 2$ gives $1 = -B$, so $B = -1$. Then

$I = \int \frac{dt}{1 - t} - \int \frac{dt}{2 - t}.$

Integrating

Dividing by the derivative of each linear term,

$I = -\log|1 - t| + \log|2 - t| + C = \log\left|\frac{2 - t}{1 - t}\right| + C.$

Substituting $t = \sin x$,

$I = \log\left|\frac{2 - \sin x}{1 - \sin x}\right| + C.$

An integral in $x^2$ split into inverse tangents

We want

$I = \int \frac{x^2}{(x^2 + 1)(x^2 + 4)} \, dx.$

Here there is no $x \, dx$ in the numerator, so we put $x^2 = t$ only to do the partial fractions, then re-substitute before integrating:

$\frac{t}{(t + 1)(t + 4)} = \frac{A}{t + 1} + \frac{B}{t + 4}.$

From $t = A(t + 4) + B(t + 1)$, putting $t = -1$ gives $-1 = 3A$, so $A = -\tfrac{1}{3}$, and putting $t = -4$ gives $-4 = -3B$, so $B = \tfrac{4}{3}$. Re-substituting $t = x^2$,

$\frac{x^2}{(x^2 + 1)(x^2 + 4)} = \frac{-\tfrac{1}{3}}{x^2 + 1} + \frac{\tfrac{4}{3}}{x^2 + 4}.$

Integrating

Using $\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$,

$I = -\frac{1}{3}\tan^{-1} x + \frac{4}{3} \cdot \frac{1}{2}\tan^{-1}\frac{x}{2} + C = -\frac{1}{3}\tan^{-1} x + \frac{2}{3}\tan^{-1}\frac{x}{2} + C.$

Question 18: dividing first, then partial fractions

We want

$I = \int \frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)} \, dx.$

Put $x^2 = t$ to simplify the ratio:

$\frac{(t + 1)(t + 2)}{(t + 3)(t + 4)} = \frac{t^2 + 3t + 2}{t^2 + 7t + 12}.$

The numerator and denominator have the same degree, so divide first:

$\frac{t^2 + 3t + 2}{t^2 + 7t + 12} = 1 + \frac{-4t - 10}{(t + 3)(t + 4)}.$

Partial fractions

Split $\dfrac{4t + 10}{(t + 3)(t + 4)} = \dfrac{A}{t + 3} + \dfrac{B}{t + 4}$, so $4t + 10 = A(t + 4) + B(t + 3)$. Putting $t = -4$ gives $-6 = -B$, so $B = 6$, and putting $t = -3$ gives $-2 = A$. Therefore

$\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)} = 1 + \frac{2}{x^2 + 3} - \frac{6}{x^2 + 4}.$

Integrating

Using the standard inverse tangent integral with $a = \sqrt{3}$ and $a = 2$,

$I = x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - 3\tan^{-1}\frac{x}{2} + C.$

Key takeaways

• Substituting $\sin x = y$ turns a trigonometric integral into a rational one; a repeated factor $(y - 2)^2$ needs both $\dfrac{A}{y - 2}$ and $\dfrac{B}{(y - 2)^2}$ terms.
• For an integrand in $x^2$ with no $x \, dx$ on top, substitute $x^2 = t$ just to find the partial fractions, then put $t = x^2$ back before integrating into inverse tangents.
• When the numerator and denominator have equal degree, divide first, then split the remainder into partial fractions.