Trigonometric Substitution
The Essentials
Trigonometric Substitution utilizes these two trigonometric identities:
\[
1=sin^2(x)+cos^2(x)
\]
\[
sec^2(x)=1+tan^2(x)
\]
These identities allow us to simplify complex integrals that contain variations of \( \sqrt{x^2+a^2} \), \( \sqrt{x^2-a^2} \), and \( \sqrt{a^2-x^2} \).
Example
We have an integral shown as: \( \int_0^\frac{\pi}{4}\frac{1}{\sqrt{9+x^2}}dx \)
Because we know that \( sec^2(\theta)=tan^2(\theta)+1 \), we can substitute \( x=3tan(\theta) \) and \( dx=3sec^2(\theta) \) to have an integral of:
\[
\int_0^{2\pi}\frac{1}{\sqrt{9+9tan^2(\theta}}3sec^2(\theta)d\theta
\]
Next we can remove the square root for a more simple equation.
\[
\int_0^{2\pi}\frac{3sec^2(\theta)}{3sec(\theta)}d\theta
\]
\[
\int_0^{2\pi}(sec(\theta))d\theta
\]
Then we find the antiderivative of \( sec(\theta) \) to be evaluated at \( 2\pi \) and 0.
\[
ln|sec(\theta)=tan(\theta)||_0^{2\pi}
\]
Answer = 0.6156
Practice
Evaluate these expressions:
- \( \int_0^\pi\frac{1}{\sqrt{16-x^2}}dx \)
- \( \int_0^\frac{\pi}{4}\frac{1}{\sqrt{x^2-81}}dx \)
Solution:
- \( -\pi \)
- 1.4841