Solving $\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx$ such that $(1\lt x\lt5)$ using trigonometric substitutions and pythagorean identities

Question

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?

Answer 1

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

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

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

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

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

Answer 2

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: $$\color{red}{\int \frac{dx}{\sqrt{x^2+a^2}}=\arcsin \frac{x}{a} + c}$$

$$$$

Answer 3

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.

Answer 4

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 $d\nu=\frac{dt}{2}$

$$\int\frac{d\nu}{\sqrt{1-\nu ^2}}=\arcsin(\nu)+\mathcal C=\color{red}{\arcsin\left(\frac{x-3}{2}\right)+\mathcal C}$$

Answer 5

HINTS:

• Substitute $x=3+2\sin\theta$
• Recall that $1-\sin^2\theta \equiv \cos^2\theta$