# How do you integrate int (4x)/sqrt(x^2-14x+45)dx using trigonometric substitution?

Feb 29, 2016

$4 \sqrt{{\left(\frac{x - 7}{2}\right)}^{2} + 1} + 7 {\cosh}^{-} 1 \left(\frac{x - 7}{2}\right) + C$

#### Explanation:

We can begin by taking ${x}^{2} - 14 x + 45$ an re writing it in in completed square/ vertex form giving us:

${\left(x - 7\right)}^{2} - 4$ Putting this into the integral will give:

$\int \frac{4 x}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$

We can factor the $4$ out to the front then play with the numerator a little and we get:

$= 4 \int \frac{x - 7 + 7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$
$= 4 \int \frac{x - 7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} + \frac{7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$

Now lets consider the substitution (we'll have to use a hyperbolic function, not a trig function): $2 \cosh \left(u\right) = x - 7$
$2 \sinh \left(u\right) \mathrm{du} = \mathrm{dx}$

Now putting this substitution in we get:

$4 \int \left(\frac{2 \cosh \left(u\right)}{\sqrt{4 {\cosh}^{2} \left(u\right) - 4}} + \frac{7}{\sqrt{4 {\cosh}^{2} \left(u\right) - 4}}\right) \sinh \left(u\right) \mathrm{du}$

Simplify this a little by factoring the $4$ from the square root:

$= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sqrt{{\cosh}^{2} \left(u\right) - 1}} + \frac{7 \sinh \left(u\right)}{2 \sqrt{{\cosh}^{2} \left(u\right) - 1}} \mathrm{du}$

Now at this point we can use the identity: ${\cosh}^{2} \left(u\right) - 1 = {\sinh}^{2} \left(u\right)$ and then do a bit of cancelling.

That will give us:

$= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sqrt{{\sinh}^{2} \left(u\right)}} + \frac{7 \sinh \left(u\right)}{2 \sqrt{{\sinh}^{2} \left(u\right)}} \mathrm{du}$

$= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sinh} \left(u\right) + \frac{7 \sinh \left(u\right)}{2 \sinh \left(u\right)} \mathrm{du}$

$= 4 \int \cosh \left(u\right) + \frac{7}{2} \mathrm{du}$

Now evaluate the integral:

$4 \left(\sinh \left(u\right) + \frac{7}{2} u\right) + C = 4 \sinh \left(u\right) + 14 u + C$

Now we can use the identity that we used earlier to convert $\sinh$ into $\cosh$:

$= 4 \left(\sqrt{{\cosh}^{2} \left(u\right) + 1} + \frac{7}{2} u\right) + C$

Now reverse the substitution and we get:

$4 \sqrt{{\left(\frac{x - 7}{2}\right)}^{2} + 1} + 7 {\cosh}^{-} 1 \left(\frac{x - 7}{2}\right) + C$

At this point you can re write the function in terms of its exponential and logarithmic definitions but I believe this maybe satisfactory>