How do you integrate #int 1/sqrt(3x-12sqrtx+40) # using trigonometric substitution?

Question

Emily Ammerman · Accepted Answer

#\frac{2}{3}(3x-12\sqrtx+40)^{1/2}+4/\sqrt3\sinh^{-1}(\frac{\sqrt{3x}-2\sqrt3}{\sqrt{28}})+C# Let #\sqrtx=t\implies \frac{dx}{2\sqrtx}=dt\ \ or\ \ dx=2t\dt# # herefore \int \frac{dx}{\sqrt{3x-12\sqrtx+40}}# #=\int \frac{2tdt}{\sqrt{3t^2-12t+40}}# #=\frac{1}{3}\int \frac{((6t-12)+12)dt}{\sqrt{3t^2-12t+40}}# #=\frac{1}{3}\int \frac{(6t-12)dt}{\sqrt{3t^2-12t+40}}+12/3\int \frac{dt}{\sqrt{3t^2-12t+40}}# #=\frac{1}{3}\int \frac{d(3t^2-12t+40)}{\sqrt{3t^2-12t+40}}+4\int \frac{dt}{\sqrt3\sqrt{t^2-4t+40/3}}# #=\frac{1}{3}\int \frac{d(3t^2-12t+40)}{(3t^2-12t+40)^{1/2}}+4/\sqrt3\int \frac{dt}{\sqrt{(t-2)^2+28/3}}# #=\frac{1}{3} \frac{(3t^2-12t+40)^{-1/2+1}}{-1/2+1}+4/\sqrt3\int \frac{d(t-2)}{\sqrt{(t-2)^2+(\sqrt{28/3})^2}}# #=\frac{2}{3}(3t^2-12t+40)^{1/2}+4/\sqrt3\sinh^{-1}(\frac{t-2}{\sqrt{28/3}})+C# #=\frac{2}{3}(3x-12\sqrtx+40)^{1/2}+4/\sqrt3\sinh^{-1}(\frac{\sqrt{3x}-2\sqrt3}{\sqrt{28}})+C#

Isabella Ackerman · Answer

undefined To integrate $ \int \frac{1}{\sqrt{3x - 12\sqrt{x} + 40}} $ using trigonometric substitution, we start by making a substitution to simplify the integral. Let's let:

$$ x = 4(\sin^2 \theta + \cos^2 \theta) $$

From this, we can find $ dx $:

$$ dx = 8\sin \theta \cos \theta \, d\theta $$

Now, we replace $ x $ and $ dx $ in the integral, and also express the expression under the square root in terms of $ \theta $:

$$ \frac{1}{\sqrt{3(4(\sin^2 \theta + \cos^2 \theta)) - 12\sqrt{4(\sin^2 \theta + \cos^2 \theta)} + 40}} $$

Simplify this expression:

$$ \frac{1}{\sqrt{12 - 24(\sqrt{\sin^2 \theta + \cos^2 \theta}) + 40}} $$

$$ = \frac{1}{\sqrt{52 - 24}} = \frac{1}{\sqrt{28}} = \frac{1}{2\sqrt{7}} $$

Now, the integral becomes:

$$ \int \frac{1}{2\sqrt{7}} \cdot 8\sin \theta \cos \theta \, d\theta $$

Simplify and rewrite:

$$ = \frac{4}{\sqrt{7}} \int \sin \theta \cos \theta \, d\theta $$

Using the identity $ \sin(2\theta) = 2\sin \theta \cos \theta $, we rewrite the integral:

$$ = \frac{4}{\sqrt{7}} \int \frac{1}{2} \sin(2\theta) \, d\theta $$

Now, integrate $ \sin(2\theta) $:

$$ = \frac{4}{\sqrt{7}} \cdot \frac{-\cos(2\theta)}{2} + C $$

Substitute back $ x $ using the original substitution:

$$ = \frac{-2\cos(2\theta)}{\sqrt{7}} + C $$

Now, we need to express $ \cos(2\theta) $ back in terms of $ x $. Since $ \cos(2\theta) = 1 - 2\sin^2 \theta $, and $ x = 4(\sin^2 \theta + \cos^2 \theta) $, we have:

$$ \cos(2\theta) = 1 - \frac{x}{2} $$

Substitute this back into the integral expression:

$$ = \frac{-2(1 - \frac{x}{2})}{\sqrt{7}} + C $$

Simplify:

$$ = \frac{-2 + x}{\sqrt{7}} + C $$

Therefore, the integral of $ \frac{1}{\sqrt{3x - 12\sqrt{x} + 40}} $ using trigonometric substitution is $ \frac{-2 + x}{\sqrt{7}} + C $.

Christopher Agresta · Answer

undefined To integrate $ \int \frac{1}{\sqrt{3x - 12\sqrt{x} + 40}} $ using trigonometric substitution, you can let $ x = (\frac{16}{9}) \sec^2(\theta) $. Then perform the substitution, simplify the expression, and integrate with respect to $ \theta $.