How do you integrate #int 1/sqrt(4x^2-12x+9) # using trigonometric substitution?

Question

Cameron Alexander · Accepted Answer

#int(1/(2x-3))dx=1/2ln(2x-3)+C# denominator is a perfect square #((2x)^2-2*2x*3+3^2)=(2x-3)^2# hence integral #=int(1/(2x-3))dx=1/2ln(2x-3)+C#

Emilia Aguilar · Answer

undefined To integrate $ \frac{1}{\sqrt{4x^2 - 12x + 9}} $ using trigonometric substitution, we first complete the square in the denominator. This gives us:

$$ \frac{1}{\sqrt{4(x^2 - 3x + \frac{9}{4})}} $$

$$ = \frac{1}{\sqrt{4\left((x - \frac{3}{2})^2 - \left(\frac{3}{2}\right)^2\right)}} $$

$$ = \frac{1}{\sqrt{4\left((x - \frac{3}{2})^2 - \frac{9}{4}\right)}} $$

$$ = \frac{1}{\sqrt{4}\sqrt{(x - \frac{3}{2})^2 - \frac{9}{4}}} $$

$$ = \frac{1}{2\sqrt{(x - \frac{3}{2})^2 - \frac{9}{4}}} $$

Now, let $ x - \frac{3}{2} = \frac{3}{2} \sec(\theta) $. Thus, $ dx = 3 \sec(\theta) \tan(\theta) \, d\theta $.

Substituting these into the integral, we get:

$$ \int \frac{1}{2\sqrt{(\frac{3}{2}\sec(\theta))^2 - \frac{9}{4}}} \cdot 3\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \frac{3}{2} \int \frac{\sec(\theta)\tan(\theta)}{\sqrt{(\frac{9}{4}\sec^2(\theta) - \frac{9}{4}})} \, d\theta $$

$$ = \frac{3}{2} \int \frac{\sec(\theta)\tan(\theta)}{\sqrt{\frac{9}{4}(\sec^2(\theta) - 1)}} \, d\theta $$

$$ = \frac{3}{2} \int \frac{\sec(\theta)\tan(\theta)}{\frac{3}{2}\tan(\theta)} \, d\theta $$

$$ = \int d\theta $$

Integrating $ d\theta $ with respect to $ \theta $ gives $ \theta + C $.

Now, we need to express the result in terms of $ x $. Recall that $ x - \frac{3}{2} = \frac{3}{2}\sec(\theta) $. Solving for $ \sec(\theta) $ gives $ \sec(\theta) = \frac{x - \frac{3}{2}}{\frac{3}{2}} = \frac{2x - 3}{3} $.

Using the identity $ \sec^2(\theta) = 1 + \tan^2(\theta) $, we find that $ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{2x - 3}{3}\right)^2 - 1} = \sqrt{\frac{4x^2 - 12x + 9}{9} - 1} = \sqrt{\frac{4x^2 - 12x + 9 - 9}{9}} = \sqrt{\frac{4x^2 - 12x}{9}} = \frac{\sqrt{4x^2 - 12x}}{3} $.

Therefore, the final result is $ \theta + C = \arccos\left(\frac{2x - 3}{3}\right) + C $.