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?

$$\int\sqrt{\frac{1}{-x^2+6x-5}}\space\text{d}x=\int\frac{1}{\sqrt{4-(x-3)^2}}\space\text{d}x=$$

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

$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\arcsin\left(s\right)+\text{C}=\arcsin\left(\frac{u}{2}\right)+\text{C}=\arcsin\left(\frac{x-3}{2}\right)+\text{C}$$
