How do you integrate #int 1/sqrt(16-x^2)# by trigonometric substitution?

Question

Christopher Allers · Accepted Answer

#=sin^(-1)(1/4x)+C# #int1/(sqrt(16-x^2))dx# #int1/(sqrt(16(1-x^2/16)))dx# #int1/(4sqrt(1-(x/4)^2))dx# #1/4int1/(sqrt(1-(x/4)^2))dx# Let #u = 1/4x# #(du)/(dx)=1/4# #4du=dx# #1/4int1/sqrt(1-u^2)4du# #int1/sqrt(1-u^2)du# #=sin^(-1)u+C# Substitute #u# back in: #=sin^(-1)(1/4x)+C#

Evelyn Agard · Answer

undefined To integrate $ \int \frac{1}{\sqrt{16-x^2}} $ by trigonometric substitution, let $ x = 4\sin(\theta) $, where $ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $. Then, $ dx = 4\cos(\theta) d\theta $. Substitute these into the integral and simplify. This yields $ \int \frac{1}{\sqrt{16-(4\sin(\theta))^2}} \cdot 4\cos(\theta) d\theta $. Simplifying under the square root gives $ \int \frac{1}{\sqrt{16-16\sin^2(\theta)}} \cdot 4\cos(\theta) d\theta = \int \frac{1}{\sqrt{16(1-\sin^2(\theta))}} \cdot 4\cos(\theta) d\theta = \int \frac{4\cos(\theta)}{4\cos(\theta)} d\theta = \int d\theta $. Integrate $ d\theta $ to get $ \theta + C $, where $ C $ is the constant of integration. Finally, substitute back $ x = 4\sin(\theta) $ to get the final answer in terms of $ x $.