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

Thus:

Integrating sine gives:

Write this in terms of sine:

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To integrate ( \int \frac{x}{\sqrt{16-x^2}} ) using trigonometric substitution, you can use the substitution ( x = 4 \sin(\theta) ), which leads to ( dx = 4 \cos(\theta) d\theta ). Then, ( \sqrt{16-x^2} ) becomes ( 4 \cos(\theta) ). Substituting these into the integral gives ( \int \frac{4 \sin(\theta)}{4 \cos(\theta)} \cdot 4 \cos(\theta) d\theta ). Simplifying, you get ( \int \sin(\theta) d\theta ), which is straightforward to integrate. After integrating, you substitute back ( \theta ) with ( \arcsin(x/4) ) to obtain the final result.

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