Solving $\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx$ such that $(1\lt x\lt5)$ using trigonometric substitutions and pythagorean identities First post so I'll get right to the question;
$$\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx,\qquad(1\lt x\lt5)$$
To begin with I completed the square which yields: $$-(x-3)^{2}+4$$
substituting this completed square into the integrand yields:
$$\int {\frac{1}{(-(x-3)^{2}+4)^{1/2}}}\,\,dx$$
solving by substituting the trig substitution:
let $$x = 3+2\tan u$$
hence: $$\frac{dx}{du} = 2\sec^{2}u$$
thus:
$$dx = (2\sec^{2}u) du$$
Rearranging $$x = 3+2\tan u$$
yields $$u=\tan^{-1}(\frac{x-3}{2})$$
Now rewriting the integral using the value of x yields:
$$\int {\frac{1}{(-(3+2\tan u-3)^{2}+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which yields:$$\int {\frac{1}{(-(2\tan u)^{2}+4)^{1/2}}}\,\,(2\sec^{2}u) du$$
which is the same as saying: $$\int {\frac{1}{(-4\tan^{2}u+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which can be rearranged to give: $$\int {\frac{1}{(4-4\tan^{2}u)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
Applying the constant multiple rule to the integral yields:
$$\frac 12\int {\frac{1}{(1-\tan^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
Now we know that pythagorean identity: $$\tan^{2}u=\sec^{2}u-1$$
so substituting I end up with an indefinite integral that looks like this:
$$\frac 12\int {\frac{1}{(1-\sec^{2}u+1)^{1/2}}}\,\,\,(\sec^{2}u) du$$
which gives:
$$\frac 12\int {\frac{1}{(2-\sec^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
However this is obviously wrong because I think it is unsolvable. So where did I veer from the right path?
 A: $$\int\sqrt{\frac{1}{-x^2+6x-5}}\space\text{d}x=\int\frac{1}{\sqrt{4-(x-3)^2}}\space\text{d}x=$$

Substitute $u=x-3$ and $\text{d}u=\text{d}x$:

$$\int\frac{1}{\sqrt{4-u^2}}\space\text{d}u=\int\frac{1}{2\sqrt{1-\frac{u^2}{4}}}\space\text{d}u=\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}\space\text{d}u=$$

Substitute $s=\frac{u}{2}$ and $\text{d}s=\frac{1}{2}\space\text{d}u$:

$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\arcsin\left(s\right)+\text{C}=\arcsin\left(\frac{u}{2}\right)+\text{C}=\arcsin\left(\frac{x-3}{2}\right)+\text{C}$$
A: Hint: 
Are you aware about the integral 
$$\int \frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\frac{t}{a}$$
Here, set $t=x-3$, $a=2$
So your integral becomes $\cdots$ ?
Edit : This is in response to OP's edited question, instead of substituting $t=3+2\tan u$, substitute $$t=3+2\sin u$$ and do the required transformations.
Also, proceeding from your approach, I say that writing $\sec x=\frac{1}{\cos x}$ and then use $$2\cos^2-1=\cos 2x$$
Will probably do the job.
A: You are on the right track. You should proceed as follows:
$$\begin{align} &\int {\left(\frac{1}{-x^{2}+6x-5}\right)}^{1/2} dx
\\ &=\int {\left(\frac{1}{4-x^{2}+6x-9}\right)}^{1/2} dx
\\ &=\int \frac{dx}{\sqrt{2^2-(x-3)^2}} 
\\ &=\int \frac{d(x-3)}{\sqrt{2^2-(x-3)^2}} 
\\ &=\color{blue}{\arcsin \frac{x-3}{2} + c}\end{align}$$ where $c$ is a constant of integration
The last step is according to integration formulae:
$$\color{red}{\int \frac{dx}{\sqrt{x^2+a^2}}=\arcsin \frac{x}{a} + c}$$
$$$$
A: You can continue from here:
$$\int\frac{dx}{\sqrt{4-(x-3)^2}}$$
Set $t=x-3$ and $dt=dx$
$$=\int\frac{dt}{\sqrt{4-t^2}}=\int\frac{dt}{2\sqrt{1-t^2/4}}$$
Set $\nu=t/2$ and $d\nu=\frac{dt}{2}$
$$\int\frac{d\nu}{\sqrt{1-\nu ^2}}=\arcsin(\nu)+\mathcal C=\color{red}{\arcsin\left(\frac{x-3}{2}\right)+\mathcal C}$$
A: HINTS:


*

*Substitute $x=3+2\sin\theta$

*Recall that $1-\sin^2\theta \equiv \cos^2\theta$

