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

Integration by Trig Substitution: Example 1

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

For this example, we will use the following integral which has been chosen because it requires this technique. 🙂

\LARGE \mathrm{\int} \normalsize \mathrm{ \dfrac{1}{\sqrt{9 - x^2}} dx}

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

 

\mathrm{Because \ it \ looks \ like \ this\:} \\ \ \\ \Large \mathrm{\frac{1}{\sqrt{a^2 - x^2}}}
\mathrm{We \ will \ use \ these \ substitutions} \\ \ \\ \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}}

Step 3: Make the substitution & simplify

At this point, we will need to use the trig identities to simplify it into an easy-to-integrate form:

\LARGE \mathrm{\int} \normalsize \mathrm{ \dfrac{1}{\sqrt{9 - x^2}} dx =} \LARGE \mathrm{\int} \normalsize \mathrm{ \dfrac{1}{\sqrt{3^2 - x^2}} dx}

so a = 3

and

x = 3 sin(u)

dx = 3 cos(u) du

This means that under the square root we will have:

(32 − 32sin2u)

(9 − 9 sin2u)

9(1 − sin2u)

9(cos2u)

So now we can rewrite our integral:

\LARGE \mathrm{\int} \normalsize \mathrm{{\dfrac{1}{\color{blue} \sqrt{9\cos^2(u)}} \color{teal} (3 \cos(u)) du }} \\ \ \\ \LARGE \mathrm{\int} \normalsize \mathrm{{\dfrac{\color{teal} 3 \cos(u)}{\color{blue} 3\cos(u)} \color{teal} du }} \\ \ \\ \LARGE \mathrm{\int} \normalsize \mathrm{\color{teal} du }

Step 4: Integrate

\\ \ \\ \LARGE \mathrm{\int} \normalsize \mathrm{\color{teal} du = u + C }

Step 5: Convert back to the original variable

Substitute back to get the answer in terms of the variable that you started with.

Because we started with x = 9 sin(u), we need to solve for u.

Divide both sides by 9, then take the inverse sine.

 

\mathrm{\dfrac{1}{3} \cdot 3\sin(u) = \dfrac{1}{3} \cdot x} \\ \ \\ \mathrm{\sin(u) = \dfrac{x}{3}} \\ \ \\ \mathrm{u = \sin^{-1}(\dfrac{x}{3})} \\ \ \\ \mathrm{so} \\ \ \\ \LARGE \mathrm{\int} \normalsize \mathrm{ \dfrac{1}{\sqrt{9 - x^2}} dx = \sin^{-1}(\dfrac{x}{3}) + C} \\ \ \\ \mathrm{for \ -\frac{\pi}{2} \le u \le \frac{\pi}{2}}

I hope this example helped. Practice with some problems that have answers so you can check your work and it will get easier with experience.

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