How do you integrate #int 1/sqrt(9x^2-18x+13) # using trigonometric substitution?

Question

Emilia Aguilar · Accepted Answer

#sinh^-1(3/2(x-1))+C#

I can give you a solution but it uses a hyperbolic substitution as oppose to a standard trig substitution but it is still very similar.

We begin by re writing the quadratic under the square root in completed square / vertex form.

#9x^2-18x+13=9(x^2-2x)+13# #=9(x-1)^2+4#

We can re write the integral as:

#intdx/sqrt(9x^2-18x+13) = intdx/sqrt(9(x-1)^2+4)#

At this point we may like to consider the substitution:

#3(x-1) = 2sinh(u) -> 9(x-1)^2=4sinh^2(u)#

It also follows from this substitution that:

#3dx = 2cosh(u)du#

Notice it is not a standard trig function but a hyperbolic function

Now putting this into the integral we get:

#2/3intcosh(u)/sqrt(4sinh^2(u)+4)du#

We can factor #4# out of the square root on the bottom (leaving us with #2#) which we will put on the front of the integral:

#1/3intcosh(u)/sqrt(sinh^2(u)+1)du#

From the hyperbolic identity: #cosh^2(u)-sinh^2(u) =1# we can replace the expression under the square root with:

#1/3intcosh(u)/sqrt(cosh^2(u))du=#

Which simplifies to:

#1/3intcosh(u)/cosh(u)du=1/3intdu=1/3u+C#

Now reverse the substitution for #u# and we get:

#1/3sinh^-1(3/2(x-1))+C#

You can leave it here or if you want to go a bit further, re-write this in terms of the definition of the #sinh# inverse function which would give:

#1/3ln{3/2(x-1)+sqrt(9/4(x-1)^2+1) }+C#

Cameron Alexander · Answer

undefined To integrate $\int \frac{1}{\sqrt{9x^2 - 18x + 13}}$ using trigonometric substitution, we can use the substitution $x = \frac{1}{3} + \frac{2}{3} \sin(\theta)$. This substitution helps simplify the expression into a form that can be integrated more easily. Here's how it works:

Given the integral $\int \frac{1}{\sqrt{9x^2 - 18x + 13}} dx$,

1. Substitute $x = \frac{1}{3} + \frac{2}{3} \sin(\theta)$ into the integral.

2. Calculate $dx$ and substitute into the integral.

3. Simplify the integrand using trigonometric identities.

4. Integrate with respect to $\theta$.

5. Substitute back for $x$ to find the final answer.

Here are the steps in detail:

Given:
$$
x = \frac{1}{3} + \frac{2}{3} \sin(\theta)
$$
Differentiate both sides with respect to $\theta$ to find $dx$:
$$
dx = \frac{2}{3} \cos(\theta) d\theta
$$
Substitute $dx$ and $x$ into the integral:
\[
\begin{aligned}
\int \frac{1}{\sqrt{9x^2 - 18x + 13}} dx &= \int \frac{1}{\sqrt{9\left(\frac{1}{3} + \frac{2}{3}\sin(\theta)\right)^2 - 18\left(\frac{1}{3} + \frac{2}{3}\sin(\theta)\right) + 13}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{9\left(\frac{1}{3} + \frac{4}{9}\sin^2(\theta)\right) - 6\left(\frac{1}{3} + \frac{2}{3}\sin(\theta)\right) + 13}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{1 + \frac{4}{9}\sin^2(\theta)}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{1 + \frac{4}{9}(1 - \cos^2(\theta))}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{1 + \frac{4}{9} - \frac{4}{9}\cos^2(\theta)}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{\frac{13}{9} - \frac{4}{9}\cos^2(\theta)}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{\frac{13}{9}\left(1 - \frac{4}{13}\cos^2(\theta)\right)}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{\frac{13}{9}}\sqrt{1 - \frac{4}{13}\cos^2(\theta)}} \frac{2}{3}\cos(\theta) d\theta \
&= \int \frac{1}{\sqrt{\frac{13}{9}}\sqrt{1 - \left(\frac{2}{\sqrt{13}}\right)^2\cos^2(\theta)}} \frac{2}{3}\cos(\theta) d\theta \
\end{aligned}
\]

At this point, we can see that the integral can be simplified further using a trigonometric identity. We can use the substitution $u = \frac{2}{\sqrt{13}}\cos(\theta)$ to simplify the integral.

Adam Adkins · Answer

undefined To integrate $ \int \frac{1}{\sqrt{9x^2 - 18x + 13}} $ using trigonometric substitution, we first complete the square in the denominator:

$$ 9x^2 - 18x + 13 = 9(x^2 - 2x) + 13 $$

$$ = 9(x^2 - 2x + 1) + 13 - 9 $$

$$ = 9(x - 1)^2 + 4 $$

Now, let $ x - 1 = \frac{2}{3} \sec(\theta) $. Then, $ dx = \frac{2}{3} \sec(\theta) \tan(\theta) d\theta $.

Substituting into the integral:

$$ \int \frac{1}{\sqrt{9x^2 - 18x + 13}} \, dx $$

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

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

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

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

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

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

$$ = \int \frac{1}{2\sqrt{\tan^2(\theta) + 1}} \cdot \frac{4}{3} \tan(\theta) \, d\theta $$

$$ = \frac{2}{3} \int \frac{\tan(\theta)}{\sqrt{\tan^2(\theta) + 1}} \, d\theta $$

Now, using the substitution $ u = \tan(\theta) $, $ du = \sec^2(\theta) \, d\theta $:

$$ = \frac{2}{3} \int \frac{du}{\sqrt{u^2 + 1}} $$

$$ = \frac{2}{3} \sinh^{-1}(u) + C $$

$$ = \frac{2}{3} \sinh^{-1}(\tan(\theta)) + C $$

Finally, substitute back $ \theta = \sec^{-1}\left(\frac{3}{2}(x - 1)\right) $ to get the final answer in terms of $ x $.