Young woman in a thoughtful pose standing in front of a blackboard that is covered in equations

Integration techniques:

Integration by Trigonometric Substitution

Integration is the reverse of differentiation, but is quite a lot more frustrating for most people. This is the forth of a series of articles on integration techniques.

Integration by Trigonometric Substitution is used when you need to integrate a function that has a quadratic in a square root that either looks like the result of using the Pythagorean theorem or can be forced into that format by completing the square (printable version). 

Step 1: See if your integral is the correct function for this technique

Does your integral contain anything that looks like any of the following?

\\ \: \\ \Large \mathrm{\frac{1}{\sqrt{a^2 - x^2}}} \\ \: \\ \mathrm{\frac{1}{\sqrt{x^2 + a^2}}} \\ \: \\ \mathrm{\frac{1}{\sqrt{x^2 - a^2}}} \\ \: \\ \mathrm{\frac{1}{\sqrt{ax^2 + bx + c}}}

Step 2: Find u and du to substitute into your integral

 

If it looks like this \\ \: \\ \Large \mathrm{\frac{1}{\sqrt{a^2 - x^2}}}
\text{Your \color{blue}u \color{black} and \color{teal}du \color{black}will look like this:} \\ \: \\ \large \mathrm{\color{blue}x = a\:sin(u) \: \color{black}and \: \\ \color{teal} dx = a\:cos(u)du \: \\ \color{black}for \: -\frac{\pi}{2} \le u \le \frac{\pi}{2}}
\Large \mathrm{\frac{1}{\sqrt{x^2 + a^2}}}
\large \mathrm{\color{blue}x = a \tan(u) \: \color{black}and \: \\ \color{teal} dx = a\:sec^2(u)du \: \\ \color{black}for \: -\frac{\pi}{2} \lt u \lt \frac{\pi}{2}}
\Large \mathrm{\frac{1}{\sqrt{x^2 - a^2}}}
\large \mathrm{\color{blue}x = a \sec(u) \: \color{black}and \: \\ \color{teal} dx = a\:\sec(u)\tan(u) du \: \\ \color{black}for \: 0 \le u \lt \frac{\pi}{2} \: and \: \pi \le u \lt \frac{3 \pi}{2} }
\\ \: \\ \Large \mathrm{\frac{1}{\sqrt{ax^2 + bx + c}}}

Factor the quadratic polynomial so it is in one of the forms above. If the x 2 term has a coefficient, factor it out of the parentheses so
(a2x2 + b2) becomes a2(x2 + b2/a2).
You can then use the rules for one of the forms above. The a2 in front of the parentheses will move to the front of the integral sign because it is a constant that is multiplying the integrand.

This can be messy, and you may need to complete the square to convert it. (Note the shortcut method at the bottom of the linked pdf!)

Step 3: Make the substitution & simplify

At this point, you may need to use the trig identities to simplify the trig functions until you reach an easy-to-integrate form:

\large \mathrm{\sin^2x + \cos^2x = 1} \\ \: \\ \mathrm{1 + \tan^2x = \sec^2 x} \\ \: \\ \mathrm{\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y } \\ \: \\ \mathrm{\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y } \\ \: \\ \mathrm{\tan(x \pm y) = \dfrac{\tan x \pm \tan y}{1 \mp \tan x \tan y}} \\ \: \\ \mathrm{\sin(2x) = 2 \sin x \cos x} \\ \: \\ \mathrm{\cos(2x) = \cos^2(x) - \sin^2(x)} \\ \: \\ \mathrm{\sin^2(x) = \dfrac{1 - \cos(2x)}{2}} \\ \: \\ \mathrm{\cos^2(x) = \dfrac{1 + \cos(2x)}{2}} \\ \: \\ \mathrm{\sin(-x) = -\sin(x)} \\ \: \\ \mathrm{\cos(-x) = \cos(x)}

Step 4: Integrate

Some handy definitions for integration: 

\large \mathrm{\int \large \sin(x) = -\cos(x) + C} \\ \: \\ \mathrm{\int \cos(x) = \sin(x) + C} \\ \: \\ \mathrm{\int \dfrac{1}{x^2 + a^2} dx = \dfrac{1}{a} \arctan\Big(\dfrac{x}{a}\Big) + C}

Step 5: Convert back to the original variable

Substitute back to get the answer in terms of the variable that you started with. You will probably need to play with trig identities and formulas again to simplify the answer after the substitution.

 

This will make more sense when you work though and example or two. Please check out my example pages! 🙂

Learning when to use which technique takes time, and struggling a bit at first is normal. Allow plenty of time so you don’t stress yourself with a time crunch, and find some practice problems to do. Some good sources are the odd-numbered exercises in the back of each chapter of your textbook, because the answers to the odds are usually in the back of the book, or a book of practice problems like Schaum’s Outline for Calculus or the Calculus For Dummies Workbook.

If you have questions or thoughts about this or think of anything you would like to see covered, please leave a comment below!