How do you integrate #int x/sqrt(144-x^2)dx# using trigonometric substitution?

Question

Charles Arocha · Accepted Answer

Sub #x = 12sinu#. Restrict #-frac{pi}{2}<\u<=frac{pi}{2}#. Note that #cosu>=0#. #frac{dx}{du} = 12cosu# #int frac{x}{sqrt(144-x^2)} dx = int frac{12sinu}{sqrt(144-(12sinu)^2)} * frac{dx}{du} du# #= int frac{12sinu}{sqrt{144(1-sin^2u)}} * (12cosu) du# #= int frac{12sinucosu}{sqrt{cos^2u}} du# #= 12 int sinu du# #= -12 cosu + C #, where #C# is the constant of integration. #= -sqrt{144-(12sinu)^2} + C # #= -sqrt{144-x^2} + C #

Ezra Adams · Answer

undefined To integrate $ \int \frac{x}{\sqrt{144-x^2}} \, dx $ using trigonometric substitution:

1. Let $ x = 12 \sin(\theta) $, where $ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $ since $ x = 12 \sin(\theta) $ for $ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $.

2. Find $ dx $ by differentiating $ x = 12 \sin(\theta) $.

3. Substitute $ x $ and $ dx $ in terms of $ \theta $ into the integral $ \int \frac{x}{\sqrt{144-x^2}} \, dx $.

4. Simplify the integrand in terms of $ \theta $.

5. Integrate with respect to $ \theta $.

6. Finally, replace $ \theta $ with $ \sin^{-1}(\frac{x}{12}) $ to obtain the result in terms of $ x $.