<s>[INST] 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?
 [/INST] 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:
$$\color{red}{\int \frac{dx}{\sqrt{x^2+a^2}}=\arcsin \frac{x}{a} + c}$$
$$$$
</s>