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

Integration by Trig Substitution: Example 2

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

For this example, we will use the following integral:

\\ \: \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sqrt{x^2 + 10x + 26}} \: dx} \\ \: \\ \mathrm{= \LARGE \int \normalsize \dfrac{1}{\sqrt{(x^2 + 10x + 25) + 1}} \: dx} \\ \: \\ \mathrm{= \LARGE \int \normalsize \dfrac{1}{\sqrt{(x^2 + 10x + 25) + 1}} \: dx} \\ \: \\ \mathrm{= \LARGE \int \normalsize \dfrac{1}{\sqrt{(x + 5)^2 + 1}} \: dx}

So in this case, our “x” is really “x + 5” and our a is 1.
Notice the sneaky trick to force it into the x2 + a2 format:
we separated 26 into 25 + 1.

Think of math as a puzzle; you need to move pieces around in order to solve it.

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

Because it looks like this: \\ \ \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sqrt{x^2 + a^2}}dx} \\ \ \\ We have: \\ \mathrm{\color{blue}x + 5 \color{black} = \tan(u) } \\ and \\ \mathrm{\color{blue}dx \color{black} = \sec^2(u) du } \\

Step 3: Make the substitution & simplify

Now we substitute into our original equation:

Because it looks like this: \\ \ \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sqrt{x^2 + a^2}}dx} \\ \ \\ becomes \\ \ \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sqrt{\tan^2(u) + 1}}(\sec^2(u)du)} \\ \ \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sqrt{\sec^2(u)}}(\sec^2(u)du)} \\ \ \\ \mathrm{\LARGE \int \normalsize \dfrac{1}{\sec(u)}(\sec^2(u)du)} \\ \ \\ \mathrm{\LARGE \int \normalsize \sec(u)du)}

Step 4: Integrate

\mathrm{\LARGE \int \normalsize \sec(u)du) = \ln|\sec(u) + tan(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 + 5 = tan (u), we can substitute (x + 5) for tan(u).

From a trig identity, we know that sec2(u) = tan2(u) + 1,
so for sec(u), we can substitute the square root of tan2(u) + 1,
but remember that tan(u) is x + 5, so putting all of this together, we get:

\mathrm{\LARGE \int \normalsize \sec(u)du) = \ln|\sec(u) + tan(u)| + C} \\ where \\ \mathrm{\sec(u) = \sqrt{tan(u) + 1}} \\ and \\ \mathrm{\tan(u) = x + 5} \\ \\ so \\ \mathrm{\LARGE \int \normalsize \sec(u)du = \ln|\sec(u) + tan(u)| + C} \\ \mathrm{\LARGE \color{blue} \int \normalsize \sec(u)du = \ln|\sqrt{(x + 5)^2 + 1} + x + 5| + C}

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!