Tangent Substitution in Integration #
We are going to cover one of the main three types of trigonometric substitution.
You might be asking: why do we need trig for this?
After all, this looks like just a rational expression:
And you’d be right: tt is rational indeed. But with methods we’ve learned so far (like \( u \)-substitution or integration by parts), there won’t be a compatible term to make progress.
That’s where the tangent substitution comes in.
Step 1: Substitution #
We substitute:
\[ x = 4 \tan \theta \]Why 4? Because when we square this, the \( 4^2 \) matches the \( 16 \) in the denominator.
Step 2: Rewrite the Denominator #
Squaring:
\[ x^2 + 16 = (4 \tan \theta)^2 + 16 = 16 \tan^2 \theta + 16 \]Factor:
\[ 16(\tan^2 \theta + 1) \]And from the Pythagorean identity:
\[ \tan^2 \theta + 1 = \sec^2 \theta \]So the denominator becomes:
\[ 16 \sec^2 \theta \]Step 3: Adjust the Integral #
Now our integral looks like:
\[ \int \frac{1}{16 \sec^2 \theta} \, dx \]That is:
\[ \frac{1}{16} \int \cos^2 \theta \, dx \]But here’s a problem: the integral is still in terms of \( dx \), while our substitution is in terms of \( \theta \).
Step 4: Differentiate the Substitution #
From:
\[ x = 4 \tan \theta \]Differentiate:
\[ dx = 4 \sec^2 \theta \, d\theta \]Step 5: Replace \( dx \) #
Substitute into the integral:
\[ \frac{1}{16} \int \cos^2 \theta \cdot (4 \sec^2 \theta) \, d\theta \]Simplify:
\[ \frac{4}{16} \int \cos^2 \theta \cdot \sec^2 \theta \, d\theta \]Since \(\cos^2 \theta \cdot \sec^2 \theta = 1\):
\[ \frac{1}{4} \int 1 d\theta \]Step 6: Integration #
This is straightforward:
\[ \frac{1}{4} \theta + C \]Step 7: Back-Substitute #
Recall:
\[ x = 4 \tan \theta \quad \Rightarrow \quad \theta = \arctan\!\left(\frac{x}{4}\right) \]So the final answer is:
\[ \int \frac{1}{x^2 + 16} \, dx = \frac{1}{4} \arctan\!\left(\frac{x}{4}\right) + C \]Final Result:
\[ \frac{1}{4} \arctan\!\left(\frac{x}{4}\right) + C \]