Topic 2 of 7

Trig substitution and partial fractions

Step 16 · Calculus 2 topics

Integrate powers of sine and cosine, use trig substitution for square roots of quadratics, and split rational functions into partial fractions.

Trigonometric integrals

For powers of sine and cosine, use identities to set up a substitution. With an odd power, save one factor for du and turn the rest into the other function with sin²x + cos²x = 1. With only even powers, use the half-angle formulas.

\sin^{2} x = \frac{1 - \cos 2x}{2} \qquad \cos^{2} x = \frac{1 + \cos 2x}{2}

Example: An odd power

Find $\int \sin^{3} x\,dx$ .

Write sin³x as (1 − cos²x) sin x.
Let u = cos x, so du = −sin x dx.
$-\int (1 - u^{2})\,du = -u + \frac{u^{3}}{3}$ .

Answer $-\cos x + \frac{\cos^{3} x}{3} + C$

Example: An even power

Find $\int \cos^{2} x\,dx$ .

Use the half-angle formula for cos²x.
Integrate: $\frac{x}{2} + \frac{\sin 2x}{4}$ .

Answer $\frac{x}{2} + \frac{\sin 2x}{4} + C$

Trigonometric substitution

A square root of a quadratic suggests a right triangle. Substituting a trig function turns the root into a single trig term.

Root	Substitute	Identity used
$\sqrt{a^{2} - x^{2}}$	$x = a\sin\theta$	$1 - \sin^{2}\theta = \cos^{2}\theta$
$\sqrt{a^{2} + x^{2}}$	$x = a\tan\theta$	$1 + \tan^{2}\theta = \sec^{2}\theta$
$\sqrt{x^{2} - a^{2}}$	$x = a\sec\theta$	$\sec^{2}\theta - 1 = \tan^{2}\theta$

Example: A quarter circle

Find $\int_{0}^{2} \sqrt{4 - x^{2}}\,dx$ .

Let x = 2 sin θ, so dx = 2 cos θ dθ and the root is 2 cos θ.
x from 0 to 2 means θ from 0 to $\frac{\pi}{2}$ , so the integral is $\int_{0}^{\pi/2} 4\cos^{2}\theta\,d\theta$ .
With the half-angle formula that is $2\theta + \sin 2\theta$ from 0 to $\frac{\pi}{2}$ .

Answerπ: a quarter of a circle of radius 2

Example: An inverse tangent

Find $\int \frac{dx}{x^{2} + 9}$ .

Let x = 3 tan θ, so dx = 3 sec²θ dθ and x² + 9 = 9 sec²θ.
The integral becomes $\int \frac{1}{3}\,d\theta = \frac{\theta}{3}$ .

Answer $\frac{1}{3}\tan^{-1}\frac{x}{3} + C$

Partial fractions

A rational function whose bottom factors can be split into simpler fractions, one for each factor. Each piece integrates to a logarithm or a power.

Example: Two linear factors

Find $\int \frac{dx}{x^{2} - 1}$ .

Write $\frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$ , so 1 = A(x + 1) + B(x − 1).
x = 1 gives A = $\frac{1}{2}$ , and x = −1 gives B = $-\frac{1}{2}$ .

Answer $\frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| + C$

Example: A repeated factor

Split $\frac{x + 3}{(x + 1)^{2}}$ .

Try $\frac{A}{x + 1} + \frac{B}{(x + 1)^{2}}$ , so x + 3 = A(x + 1) + B.
Matching the x terms gives A = 1, and then B = 2.

Answer $\frac{1}{x + 1} + \frac{2}{(x + 1)^{2}}$

Try it yourself

Work each one on paper first, then open the answer to check.

Find $\int \sin^{2} x\,dx$ .
Show answer
$\frac{x}{2} - \frac{\sin 2x}{4} + C$
Find $\int \cos^{3} x\,dx$ .
Show answer
$\sin x - \frac{\sin^{3} x}{3} + C$
Which substitution suits $\int \frac{dx}{\sqrt{16 - x^{2}}}$ ?
Show answer
x = 4 sin θ
Find $\int_{0}^{3} \sqrt{9 - x^{2}}\,dx$ .
Show answer
$\frac{9\pi}{4}$
Split $\frac{5}{(x - 2)(x + 3)}$ into partial fractions.
Show answer
$\frac{1}{x - 2} - \frac{1}{x + 3}$
Find $\int \frac{dx}{x(x + 1)}$ .
Show answer
$\ln|x| - \ln|x + 1| + C$