- #1

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## Homework Statement

Use the method of trigonometric substitution to evaluate the following:

[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]

## Homework Equations

The only relevant equation that I could think of for this one was the trig identity:

[itex]sin^{2}\vartheta + cos^{2}\vartheta = 1[/itex]

## The Attempt at a Solution

This is what I got from trying this problem:

[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]

[itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

Then, I let [itex]x^{2} = \frac{4}{9}cos\vartheta[/itex] to satisfy the definition of cosine.

Differentiating this same function with respect to [itex]x[/itex], I got:

[itex]\frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}[/itex]

Substituting these into the original function:

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\frac{2}{3}\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\frac{-1}{27}\int{\frac{cos\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

At this point, I got stuck and didn't know what to do. I tried using a modified version of the identity above and just went on to see if I could get anywhere but failed.

Thanks