How do you integrate #int 1/sqrt(3x-12sqrtx) # using trigonometric substitution?

Question

Paisley Johnson · Accepted Answer

Use the substitution #sqrtx+2=2sectheta#.

Let #I=int1/sqrt(3x-12sqrtx)dx# Complete the square in the denominator: #I=1/sqrt3int1/sqrt((sqrtx+2)^2-4)dx# Apply the substitution #sqrtx+2=2sectheta#: #I=4/sqrt3int(sec^2theta-sectheta)d theta# Integrate term by term: #I=4/sqrt3(tantheta-ln|sectheta+tantheta|)+C# Reverse the substitution: #I=2/sqrt3sqrt((sqrtx+2)^2-4)-4/sqrt3ln|sqrtx+2+sqrt((sqrtx+2)^2-4)|+C# Simplify: #I=2/sqrt3sqrt(x+4sqrtx)-4/sqrt3ln|sqrtx+2+sqrt(x+4sqrtx)|+C#

Scarlett Allison · Answer

undefined To integrate $ \int \frac{1}{\sqrt{3x - 12\sqrt{x}}} $ using trigonometric substitution, we first perform a substitution $ x = (\frac{t}{4})^2 $. This gives us $ dx = \frac{1}{2} \frac{t}{4} dt $.

Substituting these into the integral, we get:
$$ \int \frac{1}{\sqrt{3(\frac{t}{4})^2 - 12\sqrt{(\frac{t}{4})^2}}} \cdot \frac{1}{2} \frac{t}{4} dt $$

Simplifying, we obtain:
$$ \int \frac{1}{\sqrt{\frac{3}{16}t^2 - 3t}} \cdot \frac{1}{2} \frac{t}{4} dt $$

Now, we use the trigonometric substitution $ t = 2\sqrt{3}\tan{\theta} $, which gives $ dt = 2\sqrt{3}\sec^2{\theta} d\theta $.

Substituting these, we get:
$$ \int \frac{1}{\sqrt{\frac{3}{16}(2\sqrt{3}\tan{\theta})^2 - 3(2\sqrt{3}\tan{\theta})}} \cdot \frac{1}{2} (2\sqrt{3}\tan{\theta}) 2\sqrt{3}\sec^2{\theta} d\theta $$

Simplify to get:
$$ \int \frac{1}{\sqrt{\frac{3}{4}\tan^2{\theta} - 6\tan{\theta}}} \cdot \sqrt{3}\tan{\theta} \sec^2{\theta} d\theta $$

Further simplify:
$$ \int \frac{1}{\sqrt{3\tan^2{\theta} - 6\tan{\theta}}} \cdot \sqrt{3}\tan{\theta} \sec^2{\theta} d\theta $$

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

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

Now, with the substitution $ u = \tan{\theta} - \sqrt{2} $, we can simplify this integral further to evaluate it.